[leetcode] 15. 3Sum @ python

最新推荐文章于 2021-04-19 21:52:06 发布

原创最新推荐文章于 2021-04-19 21:52:06 发布 · 676 阅读

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CC 4.0 BY-SA版权

本文深入探讨了经典的三数之和算法问题，提供了一种高效解决方案，通过先排序后双指针技巧，避免了重复解，实现了O(n)的时间复杂度。

原题

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]
]

解法

先将nums升序排列, 方便后续去掉重复值, 然后遍历nums, 查找在index i 时所有可能的triplet, 由于triplet组合不能重复, 所以我们从左往右查找, 对于nums[i], 先去掉重复值, 再定义左右指针, 找可能的组合, 找到组合之后, 需要在left, right再去掉重复值.
Time: O(n)
Space: O(1)

代码

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        # edge case
        if len(nums) < 3:
            return []
        res = []
        nums.sort()
        for i in range(len(nums)):
            # skip duplicate elements
            if i > 0 and nums[i] == nums[i-1]:
                continue
            l, r = i+1, len(nums)-1
            while l < r:
                s = nums[i] + nums[l] + nums[r]
                if s < 0:
                    # move l to the right
                    l += 1
                elif s > 0:
                    # move r to the left
                    r -= 1
                else:
                    res.append([nums[i], nums[l], nums[r]])
                    # skip duplicate elements
                    while l < r and nums[l] == nums[l+1]:
                        l += 1
                    while l < r and nums[r] == nums[r-1]:
                        r -= 1
                    l += 1
                    r -= 1
                    
        return res