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## Problem

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.


Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.


Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.


Constraints:

• 3 <= nums.length <= 3000
• $-10^5 <= nums[i] <= 10^5$

## Code

class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if(nums == null || nums.length < 3) return res;

Arrays.sort(nums);

for(int i = 0; i < nums.length - 2; i++){
// 去重
if(i != 0 && nums[i] == nums[i - 1]) continue;
int left = i + 1;
int right = nums.length - 1;
while(left < right){
int sum = nums[i] + nums[left] + nums[right];
if(sum == 0){
// 去重
while(left + 1 < right && nums[left + 1] == nums[left]) left++;
while(right - 1 > left && nums[right - 1] == nums[right]) right--;
left++;
right--;
} else if (sum > 0){
right--;
} else {
left++;
}
}
}

return res;
}
}