LeetCode15 Solved #59
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| int rIdx = nums.length - 1; | ||
| for(int i = 0; i < nums.length - 2; i++) { | ||
| pick = nums[i]; | ||
| target = 0 - pick; // pick + target = 0 |
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There's no need to write 0-pick. -pick is clear enough.
Test1/src/leetcode/LeetCode15.java
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| List<Integer> list = new ArrayList<Integer>(); | ||
| list.add(nums[i]); | ||
| list.add(nums[l]); | ||
| list.add(nums[r]); |
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Consider:
new ArrayList<>(Arrays.asList(nums[i], nums[l], nums[r]));
| resultAllList.add(list); | ||
| if(nums[i] == nums[l] && nums[l] == nums[r]) { | ||
| rIdx--; | ||
| } |
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I don't understand why the condition is like this. Can you explain your thought of this if-condition?
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I was also thinking that I thought wrong.
I just solved so many only zeros case by using this.
But i think this is the wrong way.
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only 3 zeros case pass this if phrase.
what i thought at first was, if I delete one zero from the main arr, maybe it will reduce finding every case.
Then for this, I would add set that have index of one zero, and if set has that index pass it.
how about this?
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I think the problem is not about the 3 zeroes cases.
What you got was a TLE.
You can click "View all" to see the full test case.
You need to think carefully about why your code got TLE.
Time Complexity : O(NM)