3 Sum
This is a follow up question to the 2 Sum problem.
Given an integer array, return all unique triplets that sum to 0.
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]Solution:
var threeSum = function (nums) {
nums.sort((a, b) => a - b)
var out = []
function findPair(curr, target) {
var left = curr + 1
var right = nums.length - 1
while (left < right) {
var currSum = nums[left] + nums[right]
if (currSum == target) {
out.push([nums[curr], nums[left], nums[right]])
left++
right--
while (left < right && nums[left] == nums[left - 1]) left++
while (left < right && nums[right] == nums[right + 1]) right--
} else if (currSum < target) {
left++
} else {
right--
}
}
}
for (var i = 0; i < nums.length; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue
findPair(i, -nums[i])
}
return out
}