15. 3Sum

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

思路分析：降维处理，3 Sum可以降维为2 Sum，简单的过滤重复为先排序，然后用双指针或者用set来过滤重复的值。

1. 排序，头尾双指针

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        res = []
        for i in range(len(nums)):
            if nums[i] > 0:
                break
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            self.twoSum(nums, res, i)

        return res
    
    def twoSum(self, nums, res, i):
        lo = i + 1
        hi = len(nums) - 1
        while lo < hi:
            if nums[lo] + nums[hi] < -nums[i]:
                lo += 1
            elif nums[lo] + nums[hi] > -nums[i]:
                hi -= 1
            else:
                res.append([nums[i], nums[lo], nums[hi]])
                lo += 1
                hi -= 1
                while lo < hi and nums[lo] == nums[lo-1]:
                    lo += 1
                while lo < hi and nums[hi] == nums[hi+1]:
                    hi -= 1

2. 排序用set记录已经处理过的重复的数据

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        res = []
        for i in range(len(nums)):
            if nums[i] > 0:
                break
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            self.twoSum(nums, res, i)

        return res
    
    def twoSum(self, nums, res, i):
        seen = set()
        lo = i+1
        while lo < len(nums):
            left = -nums[i] - nums[lo]
            if left in seen:
                res.append([nums[i], nums[lo], left])
                while lo + 1 < len(nums) and nums[lo] == nums[lo + 1]:
                    lo += 1
            seen.add(nums[lo])
            lo += 1

3. 不用排序，结果也要用set来记录，并且是要排序的元组tuple

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = set()
        dups = set()
        for i in range(len(nums)):
            if nums[i] in dups:
                continue
            dups.add(nums[i])
            self.twoSum(nums, res, i)

        return res
    
    def twoSum(self, nums, res, i):
        seen = set()
        lo = i+1
        while lo < len(nums):
            left = -nums[i] - nums[lo]
            if left in seen:
                res.add(tuple(sorted((nums[i], nums[lo], left))))
                while lo + 1 < len(nums) and nums[lo] == nums[lo + 1]:
                    lo += 1
            seen.add(nums[lo])
            lo += 1