3Sum
Difficulty: MediumTopic: ArrayTwo PointersSortingLeetcode:15.3SumGiven 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
Solution
write solution in here
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function(nums) {
nums.sort((a, b) => a - b)
const res = []
for(let i = 0; i < nums.length; i ++) {
if (nums[i - 1] === nums[i]) continue;
let l = i + 1, r = nums.length - 1
while(l < r) {
const sum = nums[i] + nums[l] + nums[r]
if (sum === 0) {
res.push([nums[i], nums[l], nums[r]])
while(nums[l] === nums[l + 1]) l ++
while(nums[r] === nums[r - 1]) r --
l ++
r --
} else if (sum > 0) {
r --
} else {
l ++
}
}
}
return res
};