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3Sum.js
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// 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.
// Example 3:
// Input: nums = [0,0,0]
// Output: [[0,0,0]]
// Explanation: The only possible triplet sums up to 0.
// Constraints:
// 3 <= nums.length <= 3000
// -105 <= nums[i] <= 105
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function (nums) {
let arr = [],
n = nums.length;
nums.sort((a, b) => a - b);
for (let i = 0; i < n - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
let j = i + 1,
k = n - 1;
while (j < k) {
let sum = nums[i] + nums[j] + nums[k];
if (sum === 0) {
arr.push([nums[i], nums[j], nums[k]]);
while (j < k && nums[j] === nums[j + 1]) j++;
while (j < k && nums[k] === nums[k - 1]) k++;
j++;
k--;
} else if (sum >= 0) k--;
else if (sum < 0) j++;
}
}
return arr;
};