LeetCode 之 Bulls and Cows-优快云博客

本文介绍了一个实现Bulls and Cows游戏的算法，包括如何根据秘密数字和猜测提供提示，涉及数字匹配和位置匹配的概念。通过实例演示了如何计算猜数字游戏中正确数字和位置的提示。

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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".

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.

很好理解就是找两个由数字构成的字符串中数相同位置也相同的个数，和数相同但位置不同的个数。特别是第二个例子，要明白第2、3个1只算为一个crow，代码如下：

class Solution {
public:
    string getHint(string secret, string guess) {
        vector<int> avector(10,0);
        vector<int> bvector(10,0);
        int a=0,b=0;
        for(int i=0;i<secret.size();i++){
            if(secret[i]==guess[i]) {a++;}
            else{
                ++avector[secret[i]-'0'];
                ++bvector[guess[i]-'0'];
            }
        }
        for(int i=0;i<10;i++){
            if(avector[i]>=bvector[i]) b+=bvector[i];
            else if(avector[i]<bvector[i]) b+=avector[i];
        }
        return std::to_string(a)+"A"+std::to_string(b)+"B";
    }
};