LeetCode-in-All
15. 3Sum
Medium
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
Solution
class Solution {
List<List<int>> threeSum(List<int> nums) {
List<List<int>> result = [];
if (nums.length < 3) {
return result; // There can't be triplets if there are less than 3 elements.
}
nums.sort(); // Sort the array to make the two-pointer approach easier.
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue; // Skip duplicates for the first element of the triplet.
}
int target = -nums[i];
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[left] + nums[right];
if (sum == target) {
result.add([nums[i], nums[left], nums[right]]);
while (left < right && nums[left] == nums[left + 1]) {
left++; // Skip duplicates for the second element of the triplet.
}
while (left < right && nums[right] == nums[right - 1]) {
right--; // Skip duplicates for the third element of the triplet.
}
left++;
right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
return result;
}
}