15. 3Sum#
Given an integer array nums, return all the unique 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: 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 unique triplets are [-1,0,1] and [-1,-1,2].
Example 2:
Input: nums = [0,1,1]
Output: []
Explanation: There are no unique triplets that sum up to 0.
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The unique triplet is [0,0,0].
Constraints:
Two Pointers#
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
vector<vector<int>> res;
for (int i = 0; i < n - 2; ++i) {
// Remove duplicates
if (i != 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1, right = n - 1, target = -nums[i];
while (left < right) {
// Remove duplicates
if (left != i + 1 && nums[left] == nums[left - 1]) {
++left;
continue;
}
int sum = nums[left] + nums[right];
if (sum < target) {
++left;
} else if (sum > target) {
--right;
} else {
res.push_back({nums[i], nums[left], nums[right]});
++left;
--right;
}
}
}
return res;
}
};