#15 3Sum——Top 100 Liked Questions-优快云博客

本文链接：https://blog.youkuaiyun.com/suixuejie/article/details/89961055

博客围绕在给定整数数组中找出所有和为零的唯一三元组展开。先尝试暴力解法，因超时未成功；后参考网上思路，排序并使用双指针辅助，仍超时，同时考虑特殊情况优化；最后又添加判断条件，使运行时间大幅缩短，性能提升。

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Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

"""

第一次：暴力解法，三层嵌套循环，时间超时，但本地运行正确

"""

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        length = len(nums)
        result = []
        for i in range(0, length):
            for j in range(i + 1, length):
                if -(nums[i] + nums[j]) in nums[j+1:]:
                    if sorted([nums[i], nums[j], -(nums[i] + nums[j])]) not in result:
                        result.append(sorted([nums[i], nums[j], -(nums[i] + nums[j])]))
        return result

"""

Time Limit Exceeded

"""

"""

第二次：参考网上思路，先对数组排序，采用两个指针辅助，只需遍历两次。设置首尾两个指针ij，求p与ij的和，如果sum>0，其中一值需要减小，此时p是固定的，所以j前移一位，若sum<0，其中一值需要增大，i后移一位，若sum=0，将pij保存下来。但是时间超时。

应该对特殊的情况进行考虑：1）对于相同的两个p，只查找一个就可以，避免了重复；2）当找到一组pij时，如果i+1与i的值是相同时，为了避免重复，i继续想后寻找下一个不同的值；3）同2）的思想

"""
class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        result = []
        nums = sorted(nums)
        for p in range(0, len(nums) - 2):
            if p > 0 and nums[p - 1] == nums[p]:                         #1）
                continue
            i, j = p + 1, len(nums) - 1
            while i < j:
                sums = nums[p] + nums[i] + nums[j]
                if sums > 0:
                    j -= 1
                elif sums < 0:
                    i += 1
                else:
                    result.append(sorted([nums[p], nums[i], nums[j]]))
                    while i < j and nums[i] == nums[i + 1]:              #2）
                        i += 1
                    while i < j and nums[j] == nums[j - 1]:              #3）
                        j -= 1
                    j -= 1
                    i += 1
        return result
"""

Runtime: 660 ms, faster than 76.34% of Python online submissions for3Sum.

Memory Usage: 15.6 MB, less than 20.42% of Python online submissions for 3Sum.

"""

"""

第三次：又加了一个判断条件，4）因为nums是从小到大排列的，如果某个p大于零了，那么后面的也不用看了，因为三个正数的和不可能为0

"""

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        result = []
        nums = sorted(nums)
        for p in range(0, len(nums) - 2):
            if nums[p] > 0:                                           #4）
                break
            if p > 0 and nums[p - 1] == nums[p]:                      #1）
                continue
            i, j = p + 1, len(nums) - 1
            while i < j:
                sums = nums[p] + nums[i] + nums[j]
                if sums > 0:
                    j -= 1
                elif sums < 0:
                    i += 1
                else:
                    result.append(sorted([nums[p], nums[i], nums[j]]))
                    while i < j and nums[i] == nums[i + 1]:          #2）
                        i += 1
                    while i < j and nums[j] == nums[j - 1]:          #3）
                        j -= 1
                    j -= 1
                    i += 1
        return result

"""

Runtime: 560 ms, faster than 90.46% of Python online submissions for3Sum.

Memory Usage: 15.5 MB, less than 20.42% of Python online submissions for 3Sum.

"""