1、自己想法，暴力

超时

难点：去重

按照题目给的答案来看只能先排序，才能给出与之一样顺序的答案。

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        numsLen = len(nums)
        target = [[0 for i in range(numsLen)] for j in range (numsLen)]
        result = []
        for i in range(numsLen):
            for j in range(i+1, numsLen):
                target[i][j] = nums[i] + nums[j]
                for k in range(j+1, numsLen):
                    flag = 1
                    if (nums[k] + target[i][j] == 0): #查重
                        for ind in range(len(result)):
                            if (nums[k] == result[ind][2]):
                                if (nums[i] == result[ind][0] or nums[i] == result[ind][1]):
                                    flag = 0
                                    break
                        if (flag == 1):
                            index = [nums[i], nums[j], nums[k]]
                            result.append(index)
        return result
                

2、官方解答 排序+双指针

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        numsLen = len(nums)
        result=[]
        for k in range(numsLen):
            if nums[k] > 0: # 之后都大于0，结束循环
                break
            if k>0 and nums[k] == nums[k-1]: # 去重
                continue
            i, j = k+1, numsLen-1
            while i < j :
                s = nums[k] + nums[i] + nums[j]
                if s < 0:
                    i += 1
                    while i < j and nums[i] == nums[i-1]: # 跳过所有与当前nums[i]相同的值
                        i += 1
                elif s > 0:
                    j -= 1
                    while i < j and nums[j] == nums[j+1]:
                        j -= 1
                else:
                    temp = [nums[k],nums[i],nums[j]]
                    result.append(temp)
                    i += 1
                    j -= 1
                    while i < j and nums[i] == nums[i-1]:
                        i += 1
                    while i < j and nums[j] == nums[j+1]:
                        j -= 1            
        return result