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
- -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;
}
}