leetcode刷题之我VS大神

最新推荐文章于 2021-06-13 19:59:01 发布

lrq8638

最新推荐文章于 2021-06-13 19:59:01 发布

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分类专栏： python

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本文深入探讨了解决三数之和问题的两种算法实现，一种是基于排序和双指针的方法，另一种则是利用哈希表进行高效查找的策略。通过对这两种方法的详细解释和对比，读者可以了解到如何在保证效率的同时避免重复解，为解决类似问题提供思路。

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]
]

我：

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        n=len(nums)
        nums.sort()
        final_list=[]
        for i in range(0, n-1): 
            l = i + 1
            r = n - 1
            x = nums[i]
            if x>0:
                break
            if (i>0 and nums[i]==nums[i-1]):
                continue
                
            while (l < r): 
                # x 是最小的

                if (x + nums[l] + nums[r] == 0):
                    final_list.append( [x, nums[l], nums[r]])
                    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
                    found = True


                # If sum of three elements is less 
                # than zero then increment in left 
                elif (x + nums[l] + nums[r] < 0): 
                    l+=1

                # if sum is greater than zero than 
                # decrement in right side 
                else: 
                    r-=1
        return final_list

    # 反思 首先习惯了sort， 然后排除重复花了很长时间，注意用while 排除多个重复，然后一旦用while还有再次check初始条件是不是被这次循环打破，非常不elegent。

大神：

    from collections import Counter
    class Solution(object):
        def threeSum(self, nums):
           
            if len(nums) <= 2:
                return []
            counter = collections.Counter(nums)
            print(counter)
            res = []
            if counter[0] >= 3:
                res.append([0, 0, 0])
            neg = []
            pos = []
            for x in counter:
                if x < 0:
                    neg.append(x)
            for x in counter:
                if x >= 0:
                    pos.append(x)
            for n in neg:
                for p in pos:
                    x = 0 - (n + p)
                    
                    if x in counter:
                        if x in {n, p} and counter[x] >= 2:
                            res.append([n,x,p])
                        if x < n:
                            res.append([n,x,p])
                        if x > p:
                            res.append([n,x,p])
            return res
            ```
            # 反思
            大神用来count来生成字典 ，还顺便去了重，字典的效率就很高了，求相反数 。