動機

複習3sum

Problem

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]]

Example 2:

Input: nums = []Output: []

Example 3:

Input: nums = [0]Output: []

 

Constraints:

  • 0 <= nums.length <= 3000
  • -105 <= nums[i] <= 105

sol

列舉不難，但是要去重複很麻煩，所以在加的時候就要看有沒有重複

反覆做2sum並一直丟掉最小的數字

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        ret = []
        nums.sort()

        for (i,n) in enumerate(nums):
            a, b = [i+1, len(nums)-1]
            while a < b:
                tmp = n+nums[a]+nums[b]
                if tmp == 0:
                    tmp = [n,nums[a],nums[b]]
                    tmp.sort()
                    if tmp not in ret:
                        ret.append(tmp)
                    b -= 1
                    a += 1
                elif tmp > 0:
                    b -= 1
                else:
                    a += 1
        return ret