LeetCode 299. Bulls and Cows

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本文介绍了一种解决Bulls and Cows游戏的算法，通过比较秘密数字和猜测数字，计算出bulls（正确位置和数值）和cows（正确数值但错误位置）的数量。算法首先检查相同位置的匹配，然后使用计数数组记录每个数字的出现次数，最后计算不同位置的匹配数量。

Description

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.

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.

Please note that both secret number and friend’s guess may contain duplicate digits.

Examples

1、
Input: secret = "1807", guess = "7810"

Output: "1A3B"

Explanation: 1 bull and 3 cows. The bull is 8, the cows are 0, 1 and 7.

2、
Input: secret = "1123", guess = "0111"

Output: "1A1B"

Explanation: The 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow.

Note: You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.

Notes

一道简单题，若secret串和guess串相同位置元素相同，则该位置计为一个bull，若不同位置的元素相同，则相同元素的两个位置计为一个cow。那么算法就是，若遇到对应位置相等，则bull计数加一。否则在“数字出现的次数”数组上做相应记录，然后在两个数组中下标为0～9的位置，取数组较小值加起来即为cows的值。

Codes

class Solution {
public:
    string getHint(string secret, string guess) {
        int len1=secret.length(),len2=guess.length();
        int nums1[10],nums2[10];
        int bulls=0,cows=0;
        for(int i=0;i<10;++i)
        {
            nums1[i]=0;
            nums2[i]=0;
        }
        for(int i=0;i<len1;++i)
        {
            if(secret[i]==guess[i])
            {
                bulls++;
            }
            else
            {
                nums1[secret[i]-'0']++;
                nums2[guess[i]-'0']++;
            }
        }
        for(int i=0;i<10;++i)
        {
            cows+=nums1[i]<nums2[i]?nums1[i]:nums2[i];
        }
        string ans;
        string b=to_string(bulls),c=to_string(cows);
        ans=b+"A"+c+"B";
        return ans;
    }
};