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3Sum.js
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3Sum.js
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// ---------------------------- Description ------------------------------------------------------
// 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
// -----------------------------------------------------------------------------------------------
const threeSum = (nums) => {
nums.sort((a, b) => a - b);
const result = [];
const n = nums.length;
for (let i = 0; i < n - 2; i++) {
// Skip duplicate values of nums[i]
if (i > 0 && nums[i] === nums[i - 1]) {
continue;
}
let left = i + 1;
let right = n - 1;
while (left < right) {
const total = nums[i] + nums[left] + nums[right];
if (total === 0) {
result.push([nums[i], nums[left], nums[right]]);
// Skip duplicate values of nums[left] and nums[right]
while (left < right && nums[left] === nums[left + 1]) {
left++;
}
while (left < right && nums[right] === nums[right - 1]) {
right--;
}
left++;
right--;
} else if (total < 0) {
left++;
} else {
right--;
}
}
}
return result;
}
// Example usage:
const nums1 = [-1, 0, 1, 2, -1, -4];
console.log(threeSum(nums1)); // Output: [[-1, -1, 2], [-1, 0, 1]]
const nums2 = [0, 1, 1];
console.log(threeSum(nums2)); // Output: []
const nums3 = [0, 0, 0];
console.log(threeSum(nums3)); // Output: [[0, 0, 0]]