Leetcode 299 Bulls and Cows 字符串处理 统计

A就是统计猜对的同位同字符的个数 B就是统计统计猜对的不同位同字符的个数

非常简单的题

 1 class Solution {
 2 public:
 3     string getHint(string secret, string guess) {
 4         int cntA = 0, cntB = 0;
 5         int sn[10] = {0}, gn[10] = {0};
 6         for(string::size_type i = 0; i < secret.size(); ++i){
 7             if(secret[i] == guess[i]) cntA++;
 8             sn[secret[i]-‘0‘]++;
 9             gn[guess[i]-‘0‘]++;
10         }
11         for(int i = 0; i< 10; ++i){
12             cntB += min(sn[i],gn[i]);
13         }
14         char s[50];
15         sprintf(s, "%dA%dB",cntA,cntB-cntA);
16         return string(s);
17     }
18 };
时间: 2024-10-13 16:19:44

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