leetcode | 15.三数之和

15.三数之和

题目描述：
给你一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 a，b，c ，使得 a + b + c = 0 ？请你找出所有和为 0 且不重复的三元组。
注意：答案中不可以包含重复的三元组。
三数之和
示例 1：

输入：nums = [-1,0,1,2,-1,-4]
输出：[[-1,-1,2],[-1,0,1]]

代码

class Solution:
    # 三重循环
    def threeSum1(self, nums):
        nums.sort()
        n = len(nums)
        result = []
        for first in range(0, n):
            if first == 0 or nums[first] != nums[first - 1]:
                for second in range(first + 1, n):
                    if second == first + 1 or nums[second] != nums[second - 1]:
                        for third in range(second + 1, n):
                            if third == second + 1 or nums[third] != nums[third - 1]:
                                if nums[first] + nums[second] + nums[third] == 0:
                                    result.append([nums[first], nums[second], nums[third]])
        return result
    
    # 排序+双指针
    def threeSum(self, nums):
        n = len(nums)
        nums.sort()
        ans = list()
        
        # 枚举 a
        for first in range(n):
            # 需要和上一次枚举的数不相同
            if first > 0 and nums[first] == nums[first - 1]:
                continue
            # c 对应的指针初始指向数组的最右端
            third = n - 1
            target = -nums[first]
            # 枚举 b
            for second in range(first + 1, n):
                # 需要和上一次枚举的数不相同
                if second > first + 1 and nums[second] == nums[second - 1]:
                    continue
                # 需要保证 b 的指针在 c 的指针的左侧
                while second < third and nums[second] + nums[third] > target:
                    third -= 1
                # 如果指针重合，随着 b 后续的增加
                # 就不会有满足 a+b+c=0 并且 b<c 的 c 了，可以退出循环
                if second == third:
                    break
                if nums[second] + nums[third] == target:
                    ans.append([nums[first], nums[second], nums[third]])
        
        return ans


if __name__ == "__main__":
    slt = Solution()
    nums = [-1, 0, 1, 2, -1, -4]
    # res = slt.threeSum(nums)
    res = slt.threeSum1(nums)  # [[-1, -1, 2], [-1, 0, 1]]
    print(res)