package cn.edu.xidian.sselab.hashtable;

import java.util.HashMap;
import java.util.Map;
import java.util.Set;

/**
 *
 * @author zhiyong wang
 * title: Bull And Cows
 * content:
 *
ou 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 friend makes a guess, you provide a hint that indicates
how many digits in said guess match your secret number exactly in both
digit and position (called "bulls") and how many digits match the secret
number but locate in the wrong position (called "cows"). Your friend
will use successive guesses and hints to eventually derive the secret
number.
 * For example:
 *
 * Secret number:  "1807"
 * Friend's guess: "7810"
 *
 * Hint: 1 bull and 3 cows. (The bull is 8, the cows are 0, 1 and 7.)
 *
 *
Write a function to return a hint according to the secret number and
friend's guess, use A to indicate the bulls and B to indicate the cows.
In the above example, your function should return "1A3B".
 *
 * Please note that both secret number and friend's guess may contain duplicate digits, for example:
 *
 * Secret number:  "1123"
 * Friend's guess: "0111"
 *
 * In this case, the 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow, and your function should return "1A1B".
 *
 * You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.
 *
 */
public class BullAndCows {

//自己没有想出来,参考大牛的解题思路
    //这里默认secret 与 guess的长度是相等的
    //这个地方求cow的值是通过新建一个数组来保存0到9这十个数字,初始化值都为0,如果在secret中,他对应的值加1,如果在guess中,他对应的值减1
    //这样如果在secret的某位的值在数组中显示小于0,说明他以前在guess中出现过,通过guess也是这样
    public String getHint(String secret, String guess){
        int[] numbers = new int[10];
        int bull = 0;
        int cow = 0;
        if(secret == null || guess == null) return null;
        if(secret.length() == 0 || guess.length() == 0)
            return "0A0B";
        for(int i=0;i<secret.length();i++){
            if(secret.charAt(i) == guess.charAt(i))
                bull++;
            else{
                if(numbers[secret.charAt(i) - '0']++ < 0) cow++;
                if(numbers[guess.charAt(i) - '0']-- > 0) cow++;
            }
        }
        return bull + "A" + cow + "B";
    }
    
}

Bull And Cows的更多相关文章

  1. [LeetCode] Bulls and Cows 公母牛游戏

    You are playing the following Bulls and Cows game with your friend: You write a 4-digit secret numbe ...

  2. LeetCode 299 Bulls and Cows

    Problem: You are playing the following Bulls and Cows game with your friend: You write down a number ...

  3. [Leetcode] Bulls and Cows

    You are playing the following Bulls and Cows game with your friend: You write a 4-digit secret numbe ...

  4. Bulls and Cows

    You are playing the following Bulls and Cows game with your friend: You write down a number and ask ...

  5. 299. Bulls and Cows

    题目: You are playing the following Bulls and Cows game with your friend: You write down a number and ...

  6. Java [Leetcode 229]Bulls and Cows

    题目描述: You are playing the following Bulls and Cows game with your friend: You write down a number an ...

  7. Poj OpenJudge 百练 2389 Bull Math

    1.Link: http://poj.org/problem?id=2389 http://bailian.openjudge.cn/practice/2389/ 2.Content: Bull Ma ...

  8. BZOJ1754: [Usaco2005 qua]Bull Math

    1754: [Usaco2005 qua]Bull Math Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 374  Solved: 227[Submit ...

  9. [LeetCode299]Bulls and Cows

    题目: You are playing the following Bulls and Cows game with your friend: You write down a number and ...

随机推荐

  1. QDomDocument类

    QDomDocument类代表了一个XML文件 QDomDocument类代表整个的XML文件.概念上讲:它是文档树的根节点,并提供了文档数据的基本访问方法. 由于元素.文本节点.注释.指令执行等等不 ...

  2. #IOS-navigation中左滑pop的三种方法

    IOS-navigation中左滑pop的三种方法 系统自带pop方法 如果我们没有对navigation中的back按钮进行自定义,我们可以直接使用系统自带的左滑pop方法.但是如果我们对back按 ...

  3. ios中图片的绘画和截图

    ios中图片的绘画和截图 CGImageCreateWithImageInRect截图和UIGraphicsGetImageFromCurrentImageContext绘画图片 使用CGImageC ...

  4. php mysqli注意问题

    今天写了这个一段代码 function ip_get_method($action , $device){ if($action != 'search'){ request_die(false,'un ...

  5. 第一章 Android体系与系统架构

    1. Dalvik 和 ART(Android Runtime) 在Dalvik中应用好比是一辆可折叠的自行车,平时是折叠的,只有骑的时候,才需要组装起来用.在ART中应用好比是一辆组装好了的自行车, ...

  6. RMQ 与 LCA-ST算法

    RMQ算法 区间求最值的算法,用区间动态规划(nlogn)预处理,查询O(1) http://blog.csdn.net/y990041769/article/details/38405063 (PO ...

  7. IIS7.5 asp+access数据库连接失败处理 64位系统

    IIS7.5 asp+access数据库连接失败处理(SRV 2008R2 x64/win7 x64) IIS7.5不支持oledb4.0驱动?把IIS运行模式设置成32位就可以了,微软没有支持出64 ...

  8. oracle-snapshot too old 示例

    一.快照太老例子:    1.创建一个很小的undo表空间,并且不自动扩展. create undo tablespace undo_small    datafile '/u01/app/oracl ...

  9. [转]MySQL数据库备份和还原的常用命令小结

    MySQL数据库备份和还原的常用命令小结,学习mysql的朋友可以参考下: 备份MySQL数据库的命令 mysqldump -hhostname -uusername -ppassword datab ...

  10. 292. Nim Game(C++)

    292. Nim Game(C++) You are playing the following Nim Game with your friend: There is a heap of stone ...