15-ThreeSum.py

# Python3
# 时间复杂度：O(N^2)  空间复杂度：O(logN)
from typing import List

class Solution: 
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        n = len(nums)
        nums.sort()
        ans = list()

        # a + b + c = 0
        # 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 后续的增加就不会存在满足 a + b + c = 0 并且 b < c 的 c, 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

def main():
    test = Solution()
    #nums = [-1,0,1,2,-1,-4]
    #nums = [0,0,0,0]
    nums = [-1,0,1,2,-1,-4,-2,-3,3,0,4]
    res = test.threeSum(nums)
    print(res)

if __name__ == "__main__":
    main()