15. 三数之和

题目链接：国际版，国内版
公司：Meta
解法：双指针

题目描述

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.

思路

本题与两数之和非常像，区别在于这次不是找两个下标不同的元素使其和为 0 ，而是要找三个下标不同的元素。既然这样，我们也可以借鉴 two sum 的思路，即：每次固定一个数组元素 nums[i]，在剩下的数组中找两数之和，使其和为 target - nums[i] 即可。为此，我们需要一个 helper function，它的作用就是求这样的两数之和。

题目中还有一些需要注意的地方是，由于初始数组是无序且包含重复元素的，因此我们要先将原数组排序，并且在进行遍历 or 更新指针的过程中要去重。

代码参考

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        res = []
        i = 0
        while i < len(nums):
            if nums[i] > 0:
                break
            tuples = self.twoSumTarget(nums, i + 1, -nums[i])
            for j, k in tuples:
                res.append([nums[i], nums[j], nums[k]])
            while i < len(nums) - 1 and nums[i] == nums[i + 1]:
                i += 1
            i += 1
        return res

    # 注意：调用这个函数之前一定要先给 nums 排序
    def twoSumTarget(self, nums: List[int], start: int, target: int) -> List[List[int]]:
        sz = len(nums)
        res = []
        # 双指针那一套操作
        lo, hi = start, sz - 1
        while lo < hi:
            s = nums[lo] + nums[hi]
            left, right = nums[lo], nums[hi]
            if s < target:
                while lo < hi and nums[lo] == left:
                    lo += 1
            elif s > target:
                while lo < hi and nums[hi] == right:
                    hi -= 1
            else:
                res.append([lo, hi])
                while lo < hi and nums[lo] == left:
                    lo += 1
                while lo < hi and nums[hi] == right:
                    hi -= 1
        return res