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] <= 1051class Solution {
2 public List<List<Integer>> threeSum(int[] nums) {
3 Set<List<Integer>> res = new HashSet<>();
4
5 for(int i=0;i<nums.length;i++){
6 Set<Integer> seen = new HashSet<>();
7
8 for(int j=i+1;j<nums.length;j++){
9 int target = -(nums[i] + nums[j]);
10
11 if(seen.contains(target)){
12 List<Integer> temp = Arrays.asList(nums[i], nums[j], target);
13 Collections.sort(temp);
14 res.add(temp);
15 }
16
17 seen.add(nums[j]);
18 }
19 }
20
21 return new ArrayList<>(res);
22 }
23}