-
Notifications
You must be signed in to change notification settings - Fork 1
Expand file tree
/
Copy pathjava_2_2_2.java
More file actions
83 lines (77 loc) · 2.74 KB
/
java_2_2_2.java
File metadata and controls
83 lines (77 loc) · 2.74 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
package chapter2;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
/**
* 寻找和为定值的两个数
* <p>
* 输入一个整数数组和一个整数,在数组中查找一对数,满足他们的和正好是输入的那个整数。
* 不可以重复利用数组中同样的元素
*/
public class java_2_2_2 {
//三数之和为0的话,a+b+c=0 => a+b=-c
public static void main(String[] argv) {
int[] nums = {7, 9, 10, 5, 13, 1, 2, 15, 3, 4, 6, 11, 12, 8, 14};
System.out.println(Arrays.toString(twoSum1(nums, 26)));
System.out.println(Arrays.toString(twoSum2(nums, 26)));
}
//用时间换空间的做法,空间复杂度O(n),时间复杂度O(n)
//寻找两个数的下标显然可以通过key对于的value来获取
public static int[] twoSum1(int[] array, int sum) {
int[] result = new int[2];
if (array == null || array.length < 2) {
return result;
}
HashMap<Integer, List<Integer>> map = new HashMap<>();
for (int i = 0; i < array.length; i++) {
int num = array[i];
if (map.containsKey(num)) {
map.get(num).add(i);
} else {
List<Integer> indexs = new ArrayList<>();
indexs.add(i);
map.put(num, indexs);
}
}
for (int i = 0; i < array.length; i++) {
int num = array[i];
int findNum = sum - num;
if (map.containsKey(findNum)) {
List<Integer> indexs = map.get(findNum);
for (Integer index : indexs) {
if (index != i) {
//避免复用同一个元素
result[0] = num;
result[1] = findNum;
break;
}
}
}
}
return result;
}
//排序后的左右指针,排序时间复杂度O(nlogn),查找时间复杂度O(n)
//寻找两个数的下标显然也很简单,这里因为只需要找到一个结果,所以不需要去重
public static int[] twoSum2(int[] array, int sum) {
int[] result = new int[2];
if (array == null || array.length < 2) {
return result;
}
Arrays.sort(array);
int start = 0, end = array.length - 1;
while (start < end) {
int realSum = array[start] + array[end];
if (realSum < sum) {
start++;
} else if (realSum > sum) {
end--;
} else {
result[0] = array[start];
result[1] = array[end];
break;
}
}
return result;
}
}