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]]
Example 2:
Input: nums = []
Output: []
Example 3:
Input: nums = [0]
Output: []
Constraints:
- 0 <= nums.length <= 3000
- -105 <= nums[i] <= 105
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> list = new ArrayList<>();
for(int i=0;i<nums.length;i++){
// 重复元素不在计算
if(i>0&&nums[i]==nums[i-1]){
continue;
}
for(int j=i+1,k=nums.length-1;j<k;){
if(nums[i]+nums[j]+nums[k]==0){
list.add(Arrays.asList(nums[i],nums[j],nums[k]));
j++;
k--;
while(j<k&&nums[j-1]==nums[j]){
j++;
}
while(j<k&&nums[k+1]==nums[k]){
k--;
}
}else if(nums[i]+nums[j]+nums[k]<0){
j++;
}else{
k--;
}
}
}
return list;
}
}