本文主要是介绍Bulls and Cows问题及解法，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
问题描述：
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.
示例:
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.
问题分析：
统计同一位置字母相同的个数即为bull的个数，把secret和guess其他情况下的字符剩余数目分别统计，到时取两着最小值累加起来即可，也就是cow值。
过程详见代码：
class Solution {
public:string getHint(string secret, string guess) {int a = 0;int b = 0;int cache1[10];int cache2[10];fill_n( cache1, 10, 0 );fill_n( cache2, 10, 0 );for( int i = 0; i < guess.size(); ++i ){if( guess[i] == secret[i] )++a;else{cache1[secret[i]-'0']++;cache2[guess[i]-'0']++;}}for( int i = 0; i < 10; ++i ){b += min( cache1[i], cache2[i] );}return to_string(a) + "A" + to_string(b) + "B";}
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