LeetCode-Hash-299. Bulls and Cows_leetcode 299 hash-优快云博客

本文链接：https://blog.youkuaiyun.com/u011409644/article/details/54583103

本文介绍了一种解决 Bulls and Cows 游戏的算法实现。通过两次遍历，首先统计秘密数字与猜测数字在相同位置上的匹配数（bulls），并记录各自数字的出现次数；然后再次遍历比较不同位置上的匹配数（cows）。最终返回由 bulls 和 cows 组成的提示字符串。

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问题：https://leetcode.com/problems/bulls-and-cows/
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 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”.

分析：进行两次遍历。第一次，如果对应位置上secret和guess值相等，则bull值加1.同时记录两个数组中数字出现的次数，第二次遍历中，如果数字相同，存下出现的次数，它们的和就是bull+cows。

代码：

class Solution {
public:
    string getHint(string secret, string guess) {
        int l=secret.size();
        int s[10]={0};
        int g[10]={0};
        int A=0;
        int B=0;
        int sum=0;
        for(int i=0;i<l;i++){
            s[secret[i]-'0']++;
            g[guess[i]-'0']++;
            if(secret[i]==guess[i])
            A++;
        }
        for(int i=0;i<10;i++){
            sum+=min(s[i],g[i]);
        }
        B=sum-A;
        char ans[20];
        sprintf(ans, "%dA%dB", A, B);
        return ans; 
    }
};