/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function (nums) {
if (nums.length < 3) return [];
const list = [];
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length; i++) {
//nums is sorted,so it's impossible to have a sum = 0
if (nums[i] > 0) break;
// skip duplicated result without set
if (i > 0 && nums[i] === nums[i - 1]) continue;
let left = i + 1;
let right = nums.length - 1;
// for each index i
// we want to find the triplet [i, left, right] which sum to 0
while (left < right) {
// since left < right, and left > i, no need to compare i === left and i === right.
if (nums[left] + nums[right] + nums[i] === 0) {
list.push([nums[left], nums[right], nums[i]]);
// skip duplicated result without set
while (nums[left] === nums[left + 1]) {
left++;
}
left++;
// skip duplicated result without set
while (nums[right] === nums[right - 1]) {
right--;
}
right--;
continue;
} else if (nums[left] + nums[right] + nums[i] > 0) {
right--;
} else {
left++;
}
}
}
return list;
};
CPP Code:
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& A) {
sort(begin(A), end(A));
vector<vector<int>> ans;
int N = A.size();
for (int i = 0; i < N - 2; ++i) {
if (i && A[i] == A[i - 1]) continue;
int L = i + 1, R = N - 1;
while (L < R) {
int sum = A[i] + A[L] + A[R];
if (sum == 0) ans.push_back({ A[i], A[L], A[R] });
if (sum >= 0) {
--R;
while (L < R && A[R] == A[R + 1]) --R;
}
if (sum <= 0) {
++L;
while (L < R && A[L] == A[L - 1]) ++L;
}
}
}
return ans;
}
}
