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solutions/299. Bulls and Cows.md

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+# [299. Bulls and Cows](https://leetcode.com/problems/bulls-and-cows/)
+
+# 思路
+A的个数比较简单，只需要挨个位置比较即可。B的个数，需要先用一个数组记录secret中各数字出现的个数（需排除A情况），再遍历一遍看guess中的数字是否在secret中出现过，若是，则B个数加一。
+
+
+# C++
+
+
+```C++
+class Solution {
+public:
+    string getHint(string secret, string guess) {
+        int a = 0, b = 0;
+        vector<int>mp(10, 0);
+        for(int i = 0; i < secret.size(); i++){
+            if(secret[i] != guess[i]) mp[secret[i] - '0']++;
+        }
+        for(int i = 0; i < secret.size(); i++){
+            if(secret[i] == guess[i]) a++;
+            else if(mp[guess[i] - '0'] > 0){
+                b++;
+                mp[guess[i] - '0']--;
+            }
+        }
+        
+        return to_string(a) + "A" + to_string(b) + "B";
+    }
+};
+```
+也可以只遍历一遍：
+```C++
+class Solution {
+public:
+    string getHint(string secret, string guess) {
+        int a = 0, b = 0;
+        vector<int>mp1(10, 0);
+        vector<int>mp2(10, 0);
+        for(int i = 0; i < secret.size(); i++){
+            if(secret[i] != guess[i]){
+                mp1[secret[i] - '0']++;
+                mp2[guess[i] - '0']++;
+            }else{
+                a++;
+            }
+
+        }
+        for(int i = 0; i < 10; i++){
+            b += min(mp1[i], mp2[i]);
+        }
+        
+        return to_string(a) + "A" + to_string(b) + "B";
+    }
+};
+```