-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathindex.html
More file actions
2714 lines (2668 loc) · 124 KB
/
index.html
File metadata and controls
2714 lines (2668 loc) · 124 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
<!DOCTYPE html>
<html lang="zh-CN">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>LeetCode笔记</title>
<link rel="stylesheet" href="styles.css">
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/highlight.js/11.7.0/styles/github.min.css">
<link href="https://fonts.googleapis.com/css2?family=Noto+Serif:wght@400;700&family=Libre+Baskerville:wght@400;700&display=swap" rel="stylesheet">
<!-- MathJax 支持 -->
<script src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js"></script>
<script>
MathJax = {
tex: {
inlineMath: [['$', '$'], ['\\(', '\\)']],
displayMath: [['$$', '$$'], ['\\[', '\\]']],
processEscapes: true,
processEnvironments: true
},
options: {
skipHtmlTags: ['script', 'noscript', 'style', 'textarea', 'pre', 'code'],
processHtmlClass: 'tex2jax_process'
},
startup: {
pageReady: function() {
return MathJax.startup.defaultPageReady().then(function() {
console.log('MathJax initial typesetting complete');
});
}
}
};
</script>
<script src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js" id="MathJax-script" async></script>
</head>
<body>
<nav class="sidebar">
<ul id="toc">
<ul>
<li><a href="#leetcode">LeetCode 算法笔记</a></li>
<ul>
<li><a href="#_1">工具类</a></li>
<ul>
<li><a href="#_2">二分查找</a></li>
<ul>
<ul>
<li><a href="#-1">查找给定值，存在则返回索引，否则返回-1</a></li>
<li><a href="#_3">查找最后一个小于给定元素的位置（下界）</a></li>
<li><a href="#_4">查找第一个大于给定元素的位置（上界）</a></li>
</ul>
<li><a href="#_5">链表快慢指针</a></li>
<ul>
<li><a href="#_6">取链表中点</a></li>
<li><a href="#_7">找环形入口</a></li>
</ul>
<li><a href="#_8">下一个排列</a></li>
</ul>
<li><a href="#_9">滑动窗口</a></li>
<li><a href="#_10">单调栈</a></li>
<li><a href="#_11">单调队列</a></li>
<li><a href="#_12">前缀和</a></li>
</ul>
<li><a href="#_13">树</a></li>
<ul>
<li><a href="#_14">树的构造</a></li>
</ul>
<li><a href="#_15">回溯</a></li>
<li><a href="#_16">动态规划</a></li>
<ul>
<li><a href="#dp">DP 问题分类大观</a></li>
<li><a href="#dp_1">线性 DP</a></li>
<ul>
<li><a href="#_17">递推问题</a></li>
<li><a href="#_18">连续子数组问题</a></li>
<li><a href="#lis">最长递增子序列问题 LIS</a></li>
<li><a href="#_19">逆序问题</a></li>
<li><a href="#_20">双序列问题</a></li>
<li><a href="#_21">二维线性问题</a></li>
</ul>
<li><a href="#dp_2">区间DP：子区间最优解</a></li>
<ul>
<li><a href="#_22">回文串类型</a></li>
<li><a href="#_23">分割点类型</a></li>
</ul>
<li><a href="#_24">背包问题：选择与容量</a></li>
<ul>
<li><a href="#0-1">0-1 背包</a></li>
<li><a href="#_25">完全背包</a></li>
</ul>
<li><a href="#dp_3">状态机 DP</a></li>
</ul>
<li><a href="#_26">技巧</a></li>
<ul>
<ul>
<li><a href="#_27">数组元素的相互抵消运算</a></li>
<li><a href="#_28">数学技巧</a></li>
<ul>
<li><a href="#_29">矩阵旋转</a></li>
<li><a href="#_30">因式分解</a></li>
<li><a href="#_31">均值不等式</a></li>
</ul>
</ul>
</ul>
</ul>
</ul>
</ul>
</div>
</nav>
<main id="content">
<section id="leetcode"><h1 id="leetcode">LeetCode 算法笔记</h1>
<p>自用算法笔记，服务于机考；可以当作题单使用</p>
<h2 id="_1">工具类</h2>
<h3 id="_2">二分查找</h3>
<h5 id="-1">查找给定值，存在则返回索引，否则返回-1</h5>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python"># 迭代版本
def binary_search(nums, target):
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
if nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return -1
# 递归版本
def binary_search(nums, target, left, right):
if left > right:
return -1
mid = (left + right) // 2
if nums[mid] == target:
return mid
if nums[mid] < target:
return binary_search(nums, target, mid + 1, right)
else:
return binary_search(nums, target, left, mid - 1)</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">// 迭代版本
int binarySearch(vector& nums, int target) {
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] == target)
return mid;
if (nums[mid] < target)
left = mid + 1;
else
right = mid - 1;
}
return -1;
}
// 递归版本
int binarySearch(vector& nums, int target, int left, int right) {
if (left > right) {
return -1;
}
int mid = (left + right) / 2;
if (nums[mid] == target)
return mid;
if (nums[mid] < target)
return binarySearch(nums, target, mid + 1, right);
else
return binarySearch(nums, target, left, mid - 1);
}</code></pre>
</div>
</div>
</p>
<h5 id="_3">查找最后一个小于给定元素的位置（下界）</h5>
<p>找下界就是返回 <code>right</code>，<code>nums[mid] < target</code> 是否使用严格不等号等价于是否返回的是等于值的位置</p>
<p>（记：找下界，往小找，找过头了给右值）</p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python"># 迭代版本
def lower_bound(nums, target):
left = 0
right = len(nums) - 1
result = -1 # 如果所有元素都 >= target，返回-1
while left <= right:
mid = (left + right) // 2
if nums[mid] < target:
result = mid # 记录当前找到的位置
left = mid + 1
else:
right = mid - 1
return result
# 递归版本
def lower_bound(nums, target, left, right):
if left > right:
return right # 当left>right时，right就是最后一个小于target的元素
mid = (left + right) // 2
if nums[mid] < target:
return lower_bound(nums, target, mid + 1, right)
else:
return lower_bound(nums, target, left, mid - 1)</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">// 迭代版本
int lower_bound(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
int result = -1; // 如果所有元素都 >= target，返回-1
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] < target){
result = mid; // 记录当前找到的位置
left = mid + 1;
}
else
right = mid - 1;
}
return result;
}
// 递归版本
int lower_bound(vector<int>& nums, int target, int left, int right) {
if (left > right)
return right; // 当left>right时，right就是最后一个小于target的元素
int mid = (left + right) / 2;
if (nums[mid] < target)
return lower_bound(nums, target, mid + 1, right);
else
return lower_bound(nums, target, left, mid - 1);
}</code></pre>
</div>
</div>
</p>
<h5 id="_4">查找第一个大于给定元素的位置（上界）</h5>
<p>找上界就是返回 <code>left</code>，<code>nums[mid] > target</code> 是否使用严格不等号等价于是否返回的是等于值的位置</p>
<p>（记：找上界，往大找，找过头了给左值）</p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python"># 迭代版本
def upper_bound(nums, target):
left = 0
right = len(nums) - 1
result = len(nums) # 如果所有元素都 <= target，返回数组长度
while left <= right:
mid = (left + right) // 2
if nums[mid] > target:
result = mid # 记录当前找到的位置
right = mid - 1
else:
left = mid + 1
return result
# 递归版本
def upper_bound(nums, target, left, right):
if left > right:
return left # 当left>right时，left就是第一个大于target的元素
mid = (left + right) // 2
if nums[mid] > target:
return upper_bound(nums, target, mid + 1, right)
else:
return upper_bound(nums, target, left, mid - 1)</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">// 迭代版本
int upper_bound(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
int result = nums.size(); // 如果所有元素都 <= target，返回数组长度
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] > target){
result = mid; // 记录当前找到的位置
right = mid - 1;
}
else
left = mid + 1;
}
return result;
}
// 递归版本
int upper_bound(vector<int>& nums, int target, int left, int right) {
if (left > right)
return left; // 当left>right时，left就是第一个大于target的元素
int mid = (left + right) / 2;
if (nums[mid] > target)
return upper_bound(nums, target, mid + 1, right);
else
return upper_bound(nums, target, left, mid - 1);
}</code></pre>
</div>
</div>
</p>
<h4 id="_5">链表快慢指针</h4>
<h5 id="_6">取链表中点</h5>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">while fast and fast.next:
fast, slow = fast.next.next, slow.next
mid = slow.next</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">while (fast != nullptr && fast->next != nullptr) {
fast = fast->next->next;
slow = slow->next;
}
ListNode* mid = slow->next;</code></pre>
</div>
</div>
</p>
<h5 id="_7">找环形入口</h5>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python"># 此部分代码待实现</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">// 此部分代码待实现</code></pre>
</div>
</div>
</p>
<p><a href="https://leetcode.cn/problems/find-the-duplicate-number/">287. 寻找重复数 - 力扣（LeetCode）</a></p>
<h4 id="_8">下一个排列</h4>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def next_permutation(nums):
# 找到第一个降序的位置
i = len(nums) - 2
while i >= 0 and nums[i] >= nums[i + 1]:
i -= 1
if i >= 0:
# 找到比nums[i]大的最小数
j = len(nums) - 1
while j >= 0 and nums[j] <= nums[i]:
j -= 1
# 交换
nums[i], nums[j] = nums[j], nums[i]
# 反转i之后的序列
left, right = i + 1, len(nums) - 1
while left < right:
nums[left], nums[right] = nums[right], nums[left]
left += 1
right -= 1
return nums</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">#include <algorithm>
void nextPermutation(std::vector<int>& nums) {
// 找到第一个降序的位置
int i = nums.size() - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
i--;
}
if (i >= 0) {
// 找到比nums[i]大的最小数
int j = nums.size() - 1;
while (j >= 0 && nums[j] <= nums[i]) {
j--;
}
// 交换
std::swap(nums[i], nums[j]);
}
// 反转i之后的序列
std::reverse(nums.begin() + i + 1, nums.end());
}</code></pre>
</div>
</div>
</p>
<h3 id="_9">滑动窗口</h3>
<p>滑动窗口仅适用于在状态判断上单调的数组，即右扩展（例如非负数求和，长度等）</p>
<p>滑动窗口双指针的核心逻辑：</p>
<ul>
<li>右指针右移：扩展窗口，为了满足覆盖目标</li>
<li>左指针左移：缩小窗口，为了满足最小约束</li>
</ul>
<p><a href="https://leetcode.cn/problems/minimum-window-substring/description">76. 最小覆盖子串 - 力扣（LeetCode）</a></p>
<p><a href="https://leetcode.cn/problems/minimum-size-subarray-sum/description/">209. 长度最小的子数组 - 力扣（LeetCode）</a></p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def minSubArrayLen(self, target: int, nums: List[int]) -> int:
l, r = 0, 0
n = len(nums)
nums_sum=0
res = n+1
for r in range(n):
nums_sum+=nums[r]
while nums_sum>=target:
if nums_sum>=target: res=min(res, r-l+1)
nums_sum-=nums[l]
l+=1
return res if res!= n+1 else 0</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">int minSubArrayLen(int target, vector<int>& nums) {
int l=0, r=0;
int n = nums.size();
int sum = 0;
int res = n+1;
for(r=0;r<n;r++){
sum+=nums[r];
while(sum>=target){
if(sum>=target) res=min(res, r-l+1);
sum-=nums[l];
l++;
}
}
return res==n+1? 0:res;
}</code></pre>
</div>
</div>
</p>
<h3 id="_10">单调栈</h3>
<p> 在 $O(n)$ 内解决 “ 左 / 右边下一个更大 / 更小元素 ” 问题（广义：在一维数组中找第一个满足某种条件的数）：</p>
<ul>
<li>单调递增栈：从栈顶到栈底递增（口诀：<strong>递增栈找第一大</strong>）</li>
<li>满足递增：入栈时判断栈顶是否比自己<strong>小</strong>，如果是则弹出；直到<strong>栈空</strong>或栈顶<strong>不小于</strong>自身，入栈</li>
<li>当从左到右顺序遍历入栈时，入栈前的元素就是<strong>左边第一个比自己大的元素</strong></li>
<li>单调递减栈：从栈顶到栈底递减（口诀：<strong>递减栈找第一小</strong>）</li>
<li>入栈时判断栈顶是否比自己<strong>大</strong></li>
<li>如果是则弹出，直到栈顶<strong>不大于</strong>自身，入栈</li>
<li>当从左到右顺序遍历入栈时，入栈前的元素就是<strong>左边第一个比自己小的元素</strong></li>
</ul>
<p> 单调栈是非常好用的数据结构，为了避免脑子卡住：</p>
<ul>
<li>
<p>先记住口诀：减小增大（顺着的）</p>
</li>
<li>
<p>再记模板：单调栈三部曲（重在理解）</p>
</li>
</ul>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python"># 此部分代码待实现</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">while 栈非空 and 单调性不满足（这里是排除，排除不严格结果就严格）:
出栈
if 栈空: 边界处理
esle: r[i]=栈顶
无条件入栈 a[i]</code></pre>
</div>
</div>
</p>
<ul>
<li>然后理解：单调栈的精髓在于<strong>单向遮盖</strong>：对于从左向右遍历的递减栈来说，当它遇到一个值，栈中所有比它大的值都会被踢出，这个值就相当于遮盖了它左边所有比它大的值，标记出了新的左侧下界，但又保留了比自己更小的值</li>
</ul>
<p>（一般在实际应用中维护的是索引，结果序列的初试默认值为 -1，最终值为 -1 则代表没有，也就是单向极值）：</p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def next_greater(nums):
greater_stack = []
answer = [-1]*len(nums) # 初始化全 -1
for i in range(len(nums)):
while greater_stack and nums[i]>nums[greater_stack[-1]]: # 栈底在末位
index = greater_stack.pop()
answer[index] = nums[i]
greater_stack.append(i)
return answer</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">vector nextGreater(vector& nums) {
stack greaterStack;
vector answer(nums.size(), -1); // 初始化全 -1
for (int i = 0; i < nums.size(); i++) {
while (!greaterStack.empty() && nums[i] > nums[greaterStack.top()]) { // 栈底在末位
int index = greaterStack.top();
greaterStack.pop();
answer[index] = nums[i];
}
greaterStack.push(i);
}
return answer;
}</code></pre>
</div>
</div>
</p>
<p>:star:<a href="https://leetcode.cn/problems/trapping-rain-water/description">42. 接雨水 - 力扣（LeetCode）</a></p>
<p>思路：按列求，每次只关注一列（而不是块区域，这个聚焦思想很重要），一个列能够接雨水的量：取决于：<code>min(左边的最高值，右边的最高值）-当前列高</code>，于是我们只需要获得：</p>
<ul>
<li>向左看的最高列数组<code>l</code></li>
<li>向右看的最高列数组<code>r</code></li>
<li>本列的高度输入（输入）<code>a</code></li>
</ul>
<p> 就可以得到答案数组<code>res[i]=min(l[i],r[i])-a[i]</code>，对 <code>res</code> 的所有非 0 项求和即可</p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def trap(height):
n = len(height)
h = 0
l, r, res = [0] * n, [0] * n, [0] * n
for i in range(n):
l[i] = h
if height[i] > h:
h = height[i]
h = 0
for i in range(n-1, -1, -1):
r[i] = h
if height[i] > h:
h = height[i]
ans = 0
for i in range(n):
res[i] = min(l[i], r[i]) - height[i]
if res[i] > 0:
ans += res[i]
return ans</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">int trap(vector<int>& height) {
int n = height.size();
int h = 0;
vector<int> l(n), r(n), res(n);
for(int i=0;i<n;i++){
l[i] = h;
if(height[i]>h) h = height[i];
}
h = 0;
for(int i=n-1;i>=0;i--){
r[i] = h;
if(height[i]>h) h = height[i];
}
int ans = 0;
for(int i=0;i<n;i++){
res[i]=min(l[i], r[i]) - height[i];
if(res[i]>0) ans+=res[i];
}
return ans;
}</code></pre>
</div>
</div>
</p>
<p><a href="https://leetcode.cn/problems/largest-rectangle-in-histogram/description">84. 柱状图中最大的矩形 - 力扣（LeetCode）</a></p>
<p>经典单调栈题目</p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def largestRectangleArea(self, heights):
# 对于i列，矩形大小为：左边最远不小于自己的，右边最远不小于自己的
n = len(heights)
if n == 0:
return heights[0]
s = []
left, right = [0] * n, [0] * n
# 从左往右，找左边严格小,严格单调递减栈
for i in range(n):
while s and heights[s[-1]] >= heights[i]:
s.pop()
left[i] = -1 if not s else s[-1]
s.append(i)
s = []
# 从右往左，找右边严格小,严格单调递减栈
for i in range(n-1, -1, -1):
while s and heights[s[-1]] >= heights[i]:
s.pop()
right[i] = n if not s else s[-1]
s.append(i)
# 计算答案
maxa = 0
for i in range(n):
maxa = max(maxa, heights[i] * (right[i] - left[i] - 1))
return maxa</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">int largestRectangleArea(vector<int>& heights) {
// 对于i列，矩形大小为：左边最远不小于自己的，右边最远不小于自己的
int n = heights.size();
if(n==0) return heights[0];
stack<int> s;
vector<int> left(n), right(n);
// 从左往右，找左边严格小,严格单调递减栈
for(int i=0;i<n;i++){
while(!s.empty() && heights[s.top()]>=heights[i]) s.pop();
if(s.empty()) left[i] = -1;
else left[i] = s.top();
s.push(i);
}
while(!s.empty()) s.pop();
// 从右往左，找右边严格小,严格单调递减栈
for(int i=n-1;i>=0;i--){
while(!s.empty() && heights[s.top()]>=heights[i]) s.pop();
if(s.empty()) right[i] = n;
else right[i] = s.top();
s.push(i);
}
// 计算答案
int maxa = 0;
for(int i=0;i<n;i++){
maxa = max(maxa, heights[i]*(right[i]-left[i]-1));
}
return maxa;
}</code></pre>
</div>
</div>
</p>
<p><a href="https://leetcode.cn/problems/daily-temperatures/description">739. 每日温度 - 力扣（LeetCode）</a></p>
<p>单调栈经典例题，提醒一下：单调栈经常存储的是<strong>下标</strong>，而属性信息则使用下标查表</p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python"># 此部分代码待实现</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">vector<int> dailyTemperatures(vector<int>& temperatures) {
int n = temperatures.size();
vector<int> ans(n, 0);
stack<int> s; // 只存储索引，不需要结构体
for (int i = n - 1; i >= 0; i--) {
while (!s.empty() && temperatures[i] >= temperatures[s.top()]) s.pop();
if (!s.empty()) ans[i] = s.top() - i;
s.push(i);
}
return ans;
}</code></pre>
</div>
</div>
</p>
<p>四道同思路的拓展题：</p>
<p><a href="https://leetcode.cn/problems/remove-k-digits/description/">402. 移掉 K 位数字 - 力扣（LeetCode）</a></p>
<p><a href="https://leetcode.cn/problems/remove-duplicate-letters/">316. 去除重复字母 - 力扣（LeetCode）</a></p>
<p><a href="https://leetcode.cn/problems/shortest-unsorted-continuous-subarray/description/">581. 最短无序连续子数组 - 力扣（LeetCode）</a></p>
<h3 id="_11">单调队列</h3>
<p><a href="https://leetcode.cn/problems/sliding-window-maximum/description">239. 滑动窗口最大值 - 力扣（LeetCode）</a></p>
<h3 id="_12">前缀和</h3>
<p><a href="https://leetcode.cn/problems/subarray-sum-equals-k/description">560. 和为 K 的子数组 - 力扣（LeetCode）</a></p>
<p><a href="https://leetcode.cn/problems/path-sum-iii/description">437. 路径总和 III - 力扣（LeetCode）</a></p>
<h2 id="_13">树</h2>
<h3 id="_14">树的构造</h3>
<p><a href="https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/">105. 从前序与中序遍历序列构造二叉树 - 力扣（LeetCode）</a></p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
def recur(root, left, right):
if left > right: return None # 递归终止
node = TreeNode(preorder[root]) # 建立根节点
i = dic[preorder[root]] # 划分根节点、左子树、右子树
node.left = recur(root + 1, left, i - 1) # 开启左子树递归
node.right = recur(i - left + root + 1, i + 1, right) # 开启右子树递归
return node # 回溯返回根节点
dic, preorder = {}, preorder
for i in range(len(inorder)):
dic[inorder[i]] = i
return recur(0, 0, len(inorder) - 1)</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">map<int, int> index;
TreeNode* myBuildTree(const vector<int>& preorder, const vector<int>& inorder, int preorder_left, int preorder_right, int inorder_left, int inorder_right) {
if (preorder_left > preorder_right) {
return nullptr;
}
// 前序遍历中的第一个节点就是根节点
int preorder_root = preorder_left;
// 在中序遍历中定位根节点
int inorder_root = index[preorder[preorder_root]];
// 先把根节点建立出来
TreeNode* root = new TreeNode(preorder[preorder_root]);
// 得到左子树中的节点数目
int size_left_subtree = inorder_root - inorder_left;
// 递归地构造左子树，并连接到根节点
// 先序遍历中「从 左边界+1 开始的 size_left_subtree」个元素就对应了中序遍历中「从 左边界 开始到 根节点定位-1」的元素
root->left = myBuildTree(preorder, inorder, preorder_left + 1, preorder_left + size_left_subtree, inorder_left, inorder_root - 1);
// 递归地构造右子树，并连接到根节点
// 先序遍历中「从 左边界+1+左子树节点数目 开始到 右边界」的元素就对应了中序遍历中「从 根节点定位+1 到 右边界」的元素
root->right = myBuildTree(preorder, inorder, preorder_left + size_left_subtree + 1, preorder_right, inorder_root + 1, inorder_right);
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
int n = preorder.size();
for (int i = 0; i < n; ++i) {
index[inorder[i]] = i;
}
return myBuildTree(preorder, inorder, 0, n - 1, 0, n - 1);
}</code></pre>
</div>
</div>
</p>
<p><a href="https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/">106. 从中序与后序遍历序列构造二叉树 - 力扣（LeetCode）</a></p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
def helper(in_left, in_right):
# 如果这里没有节点构造二叉树了，就结束
if in_left > in_right:
return None
# 选择 post_idx 位置的元素作为当前子树根节点
val = postorder.pop()
root = TreeNode(val)
# 根据 root 所在位置分成左右两棵子树
index = idx_map[val]
# 构造右子树
root.right = helper(index + 1, in_right)
# 构造左子树
root.left = helper(in_left, index - 1)
return root
# 建立（元素，下标）键值对的哈希表
idx_map = {val:idx for idx, val in enumerate(inorder)}
return helper(0, len(inorder) - 1)</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">map<int,int> idx_map;
int post_idx = 0;
TreeNode* build(int in_left, int in_right, vector<int>& inorder, vector<int>& postorder){
if(in_left>in_right || post_idx<0) return nullptr;
int root_val = postorder[post_idx];
post_idx--;
TreeNode* root = new TreeNode(root_val);
int root_index = idx_map[root_val];
// 注意：必须是先右后左，因为是先左后右逆着来
root->right = build(root_index+1, in_right,inorder,postorder);
root->left = build(in_left, root_index-1,inorder,postorder);
return root;
}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
post_idx = postorder.size()-1;
for(int i=0;i<inorder.size();i++){
idx_map[inorder[i]]=i;
}
return build(0,inorder.size()-1,inorder,postorder);
}</code></pre>
</div>
</div>
</p>
<p><a href="https://leetcode.cn/problems/binary-tree-maximum-path-sum/description/">124. 二叉树中的最大路径和 - 力扣（LeetCode）</a></p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def maxsum(self, p):
if p == None: return 0
left = max(self.maxsum(p.left), 0)
right = max(self.maxsum(p.right), 0)
price = p.val + left + right
self.maxn = max(self.maxn, price)
return p.val + max(left, right)
def maxPathSum(self, root: Optional[TreeNode]) -> int:
# 最大和的思想往往转变为对正数的无条件合并
self.maxn = float("-inf")
self.maxsum(root)
return self.maxn</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">// 此部分代码待实现</code></pre>
</div>
</div>
</p>
<h2 id="_15">回溯</h2>
<p><a href="https://leetcode.cn/problems/non-decreasing-subsequences/description/">491. 非递减子序列 - 力扣（LeetCode）</a></p>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def findSubsequences(self, nums):
n = len(nums)
res = [] # 总的答案
path = [] # 各个答案
def dfs(start):
if len(path)>=2: res.append(path[:])
vis = set()
for i in range(start, n):
if (len(path)!=0 and nums[i]<path[-1]) or nums[i] in vis: continue
vis.add(nums[i])
path.append(nums[i])
dfs(i+1)
path.pop()
dfs(0)
return res</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">vector<vector<int>> res;
vector<int> path;
void dfs(vector<int>& nums, int start){
if(path.size()>=2) res.push_back(path);
set<int> vis;
for(int i=start;i<nums.size();i++){
if((!path.empty()&&nums[i]<path.back())||
vis.find(nums[i]) != vis.end()) continue;
vis.insert(nums[i]);
path.push_back(nums[i]);
dfs(nums, i+1);
path.pop_back();
}
}
vector<vector<int>> findSubsequences(vector<int>& nums) {
dfs(nums, 0);
return res;
}</code></pre>
</div>
</div>
</p>
<h2 id="_16">动态规划</h2>
<h3 id="dp">DP 问题分类大观</h3>
<table>
<thead>
<tr>
<th>问题特征</th>
<th>可能类型</th>
</tr>
</thead>
<tbody>
<tr>
<td>涉及数组/字符串的<strong>单个序列</strong>操作，状态与序列位置直接相关（仅依赖序列前驱）</td>
<td><strong>线性DP</strong></td>
</tr>
<tr>
<td>涉及<strong>两个序列</strong>的匹配或比较</td>
<td><strong>双序列DP</strong></td>
</tr>
<tr>
<td><strong>选择物品</strong>且有<strong>容量限制</strong>，状态定义中包含容量维度</td>
<td><strong>背包问题</strong></td>
</tr>
<tr>
<td>操作对象是<strong>区间或子序列</strong>（需枚举分割点）</td>
<td><strong>区间DP</strong></td>
</tr>
<tr>
<td>问题在<strong>树形结构</strong>上操作</td>
<td><strong>树形DP</strong></td>
</tr>
<tr>
<td>存在多个<strong>互斥状态</strong>，状态间有明确的转移规则。</td>
<td><strong>状态机DP</strong></td>
</tr>
</tbody>
</table>
<h3 id="dp_1">线性 DP</h3>
<p><strong>核心特点</strong>：状态与序列位置直接相关</p>
<p><strong>经典问题</strong>：最长递增子序列（LIS）、最大子数组和</p>
<p>🎯 <strong>状态定义套路</strong>：<code>dp[i]</code>：<strong>以第 i 个元素结尾</strong>的某种最优解</p>
<h4 id="_17">递推问题</h4>
<p><a href="https://leetcode.cn/problems/frog-jump/description/">403. 青蛙过河 - 力扣（LeetCode）</a></p>
<p><strong>问题</strong>：如果青蛙上一步跳跃了 <code>k</code> 个单位，那么它接下来的跳跃距离只能选择为 <code>k - 1</code>、<code>k</code> 或 <code>k + 1</code> 个单位。升序的 <code>stones</code> 列表指出了石头位置。</p>
<ul>
<li><strong>问题分析</strong>：</li>
<li>此处有两个需要记忆的：1、能不能跳到；2、上一步跳的距离</li>
<li>此处用倒序 <code>dp</code>是不合适的，因为现在能跳多远受到了前面的限制，而且并不是遮盖问题，是有上下界的</li>
<li>需要注意的是，石头间的距离是固定的，这意味着间隔 = 跳跃距离</li>
<li><strong>状态定义</strong>：<code>dp[i][k]</code>: 用 <code>k</code> 步跳到 <code>stones[i]</code> 的可能性</li>
<li><strong>状态转移</strong>：<code>dp[i][k] = dp[j][k] || dp[j][k+1] || dp[j][k-1], j=idx(stones[i]-k)</code></li>
<li>这里反向找索引是没必要的，因为 <code>k=stones[i]-stones[j]</code>,可以计算 k 而不是 <code>j</code> 的位置</li>
</ul>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def canCross(self, stones: List[int]) -> bool:
if stones[1]!=1:return False
n = len(stones)
dp=[[False]*(n+1) for _ in range(n)]
dp[1][1]=True
for i in range(2, n):
for j in range(1, i):
k = stones[i]-stones[j]
if k<n: dp[i][k] = dp[j][k] or dp[j][k+1] or dp[j][k-1]
for i in range(len(dp[0])):
if dp[-1][i]: return True
return False</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">// 此部分代码待实现</code></pre>
</div>
</div>
</p>
<h4 id="_18">连续子数组问题</h4>
<p>🎯 <strong>状态定义</strong>：<code>dp[i]</code>：以第 i 个元素<strong>结尾</strong>的连续子数组问题</p>
<p><strong>状态转移</strong>：<code>dp[i]=f(nums[i], g(dp[i-1],nums[i]))</code>：要么追加，要么单开，其中 <code>f</code> 是状态选择函数，<code>g</code> 是状态计算函数</p>
<p><strong>空间优化</strong>：如果只关注前驱，只保留 <code>pre</code> 即可</p>
<p><a href="https://leetcode.cn/problems/maximum-subarray/description">53. 最大子数组和 - 力扣（LeetCode）</a></p>
<p><strong>问题</strong>：找出一个具有最大和的连续非空子数组</p>
<ul>
<li><strong>问题分析</strong>：</li>
<li>连续 -> 连续子数组问题（要么追加，要么单开）</li>
<li>最大和 -> 状态转移 <code>f:max</code>，<code>g:+</code></li>
<li><strong>状态定义</strong>：<code>dp[i]</code>：以第 i 个元素结尾的最大子数组和</li>
<li><strong>状态转移方程</strong>：<code>dp[i] = max(nums[i],dp[i-1]+nums[i]);</code></li>
<li><strong>最终目标</strong>：<code>max(dp)</code></li>
<li><strong>边界条件与限制</strong>：<code>dp[0] = nums[0]</code> </li>
</ul>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def maxSubArray(self, nums: List[int]) -> int:
dp = [0]*len(nums)
dp[0] = nums[0];
for i in range(1, len(nums)):
dp[i] = max(nums[i],dp[i-1]+nums[i])
return max(dp)</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">int maxSubArray(vector<int>& nums) {
vector<int> dp(nums.size());
dp[0] = nums[0];
int maxn = dp[0];
for(int i=1;i<nums.size();i++){
dp[i] = max(nums[i],dp[i-1]+nums[i]);
if(maxn<dp[i]) maxn=dp[i];
}
return maxn;
}</code></pre>
</div>
</div>
</p>
<ul>
<li><strong>空间优化</strong>：只关注直接前驱，考虑使用前驱变量或者直接原地修改</li>
</ul>
<p>
<div class="code-container">
<div class="code-tabs">
<button class="tab-btn active" data-lang="python">Python</button>
<button class="tab-btn" data-lang="cpp">C++</button>
</div>
<div class="code-blocks">
<pre class="code-block active" data-lang="python"><code class="language-python">def maxSubArray(self, nums: List[int]) -> int:
for i in range(1, len(nums)):
nums[i] += max(nums[i - 1], 0)
return max(nums)</code></pre>
<pre class="code-block" data-lang="cpp"><code class="language-cpp">int maxSubArray(vector<int>& nums) {
int maxn = nums[0];
for(int i=1;i<nums.size();i++){
nums[i] += max(nums[i-1],0);
if(maxn<nums[i]) maxn=nums[i];
}
return maxn;
}</code></pre>
</div>
</div>
</p>
<p><a href="https://leetcode.cn/problems/maximum-product-subarray/description">152. 乘积最大子数组 - 力扣（LeetCode）</a></p>
<p><strong>问题</strong>：找出一个具有最大乘积的连续非空子数组</p>
<ul>
<li><strong>问题分析</strong>：</li>
<li>连续 -> 连续子数组问题（要么追加，要么单开）</li>
<li>最大乘积 -> 状态转移 <code>f:max</code>，<code>g:*</code></li>
<li>问题特性：对于乘法，负数会逆转结果，所以只维护最大值是不行的，还要考虑负半轴的最小值</li>
<li><strong>状态定义</strong>：<code>dp[i]</code>：以第 i 个元素结尾的最大子数组乘积</li>