Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, 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]]
3 <= nums.length <= 3000-10^5 <= nums[i] <= 10^5Sort the array first. Fix one element and use two pointers for the remaining two.
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue; // skip duplicates
int left = i + 1, right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == 0) {
result.add(List.of(nums[i], nums[left], nums[right]));
while (left < right && nums[left] == nums[left + 1]) left++;
while (left < right && nums[right] == nums[right - 1]) right--;
left++; right--;
} else if (sum < 0) left++;
else right--;
}
}
return result;
}Time: O(n²) · Space: O(1)