3sum
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.
Solution
3sum
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function (nums) {
const result = [];
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length; i++) {
if (nums[i] === nums[i - 1]) {
continue;
}
let start = i + 1;
let end = nums.length - 1;
while (start < end) {
const sums = nums[i] + nums[start] + nums[end];
if (sums > 0) {
end--;
} else if (sums < 0) {
start++;
} else {
result.push([nums[i], nums[start], nums[end]]);
while (nums[start] === nums[start + 1]) {
start++;
}
while (nums[end] === nums[end - 1]) {
end--;
}
start++;
end--;
}
}
}
return result;
};