第二天 3 Sum

作者: 业余马拉松选手 | 来源:发表于2018-08-22 01:01 被阅读8次

这道题是昨天就想好要来刷的,继续用Python,感觉通过刷题,来对这门语言有点感觉,也是个不错的方法吧

https://leetcode-cn.com/problems/3sum/description/

题目在昨天的两数之和的基础上其实难度还是增加了不少。

最初拿到这道题的思路,其实就是希望将问题“退化”到两数之和。依次遍历给定的列表,取出的值先取反,如果剩下的列表中有两数之和等于这个取反之后的值,就构成了一组解。至于去重的话,就干脆用了set,自己就没有再深入做了

为了遍历和后续处理数据方便,上来就需要把列表排个序【反正这次不需要返回在原来数组的索引值了】

接着当取出一个值后,“退化”为两数之和的时候,因为是有顺序的,那么就可以相当用两个指针,分别指向一头一尾,采取两头夹逼的方法。这里特别需要注意的时候,当找到一组解之后,还需要再继续寻找。

那么根据这个思路,我就写出了这样一段代码:

class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        ret = set()
        nums.sort()
        for k in range(len(nums)):
            i = k+1
            j = len(nums)-1
            target = -nums[k]
            while i<j:
                two_sum = nums[i] + nums[j]
                if two_sum == target:
                    ret.add((nums[k],nums[i],nums[j]))
                    i+=1
                    j-=1
                elif two_sum > target:
                    j-=1
                elif two_sum < target:
                    i+=1
        return list(ret)

结果很遗憾,TLE了
没过的case:

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开始本来认为是两侧夹逼的思路不如昨天那种字典的思路,但后来再仔细看了一下这个跑挂了的测试用例,才觉得问题应该是出现在是否需要退化到“二数之和”的部分,如果列表中两个值相等,那么前一个是否有结果,与后面一个应该是一样的,那么也就不需要循环了。
另外还有个小的潜在的优化点,就是循环的时候,只需要到倒数第三个值就可以了。
好,那么简单修改后的代码如下:

class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        ret = set()
        nums.sort()
        for k in range(len(nums)-2):
            if (k>0 and nums[k] == nums[k-1]):
                continue
            i = k+1
            j = len(nums)-1
            target = -nums[k]
            while i<j:
                two_sum = nums[i] + nums[j]
                if two_sum == target:
                    ret.add((nums[k],nums[i],nums[j]))
                    i+=1
                    j-=1
                elif two_sum > target:
                    j-=1
                elif two_sum < target:
                    i+=1
        return list(ret)

这样修改后,可以AC飘过,但也还要1秒多,看到leetcode上最好的算法只要用400ms,当然如果换成Java,“平铺直叙”的写法也只要100ms以内,这。。。。。。

好吧,其实我就是为了学习下Python而已

今天对于set的用法稍微实践了一下,另外遍历了的时候也可以使用range这种方法。

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