[LeetCode]299. Bulls and Cows 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/bulls-and-cows/description/ 题目描述: You are playing the following Bulls and Cows game with your friend: You write down…

题目如下: 解题思路:本题难度不太大,对时间复杂度也没有很高的要求.我的做法是用一个字典来保存每个字符出现的次数,用正数1记录标记secret中出现的字符,用负数1记录guess中出现的字符,这样每出现一次正负抵消,即表示出现了一次cow. 代码如下: class Solution(object): def getHint(self, secret, guess): """ :type secret: str :type guess: str :rtype: str &quo…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your >…

#-*- coding: UTF-8 -*-class Solution(object):      def getHint(self, secret, guess):          """         :type secret: str         :type guess: str         :rtype: str         """          countA,i,numList=0,-1,[[0,0] for j…

class Solution(object):    def getHint(self, secret, guess):        """        :type secret: str        :type guess: str        :rtype: str        """        countA,i,numList=0,-1,[[0,0] for i in range(10)]        while True:…

Question 299. Bulls and Cows Solution 题目大意:有一串隐藏的号码,另一个人会猜一串号码(数目相同),如果号码数字与位置都对了,给一个bull,数字对但位置不对给一个cow,注:数字对与位置对优先,一个号码不能重复判断. 思路:构造map结构,遍历实现 Java实现:实现的不漂亮,好歹能通过 public String getHint(String secret, String guess) { Map<Character, Index> map = new…

[LeetCode]738. Monotone Increasing Digits 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/monotone-increasing-digits/description/ 题目描述: Given a non-negative integer N, find th…

[LeetCode]386. Lexicographical Numbers 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/lexicographical-numbers/description/ 题目描述: Given an integer n, return 1 - n in lexicogra…

[LeetCode]Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 递归和非递归,此提比较简单.广度优先遍历即可.关键之处就在于如何保持访问深度. 下面是4种代码: im…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3,3,1]. Note: Could you optimize your algorithm to use only O(k) extra space? 思路:最简单的方法就是依照[Leetcode]Pascal's Triangle 的方式自顶向下依次求解,但会造成空间的浪费.若仅仅用一个vect…