299. Buls an Cows

ou 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.

Example 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.
Example 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.

方法1: hash table

思路：

hash secrete with unordered_map<char, int>，同时统计bulls。然后遍历guess，如果存在但是位置不在set当中 B++，并且将该字符余留cows数量–，否则跳过。

易错点

1. 第二次遍历需要跳过bulls，也就是判断不在正确位置却存在于map
2. cows–，重复的出现在guess当中并不能增加cows，必须在secret中还有余量

Complexity

O(n)

class Solution {
public:
    string getHint(string secret, string guess) {
        unordered_map<char, int> hash;
        int A = 0, B = 0;
        for (int i = 0; i < secret.size(); i++){
            if (secret[i] == guess[i]){
                A++;
            }
            else{
                ++hash[secret[i]];
            }
        }
        
        for (int i = 0; i < guess.size(); i++){
            if (secret[i] != guess[i] && hash[guess[i]]){
                B++;
                --hash[guess[i]];
            }
        }
        
        return to_string(A) + "A" + to_string(B) + "B";
    }
};

方法2:

思路:
grandyang: http://www.cnblogs.com/grandyang/p/4929139.html

这个就过于巧妙了。
我们其实可以用一次循环就搞定的，在处理不是bulls的位置时，我们看如果secret当前位置数字的映射值小于0，则表示其在guess中出现过，cows自增1，然后映射值加1，如果guess当前位置的数字的映射值大于0，则表示其在secret中出现过，cows自增1，然后映射值减1，参见代码如下：

class Solution {
public:
    string getHint(string secret, string guess) {
        int m[256] = {0}, bulls = 0, cows = 0;
        for (int i = 0; i < secret.size(); ++i) {
            if (secret[i] == guess[i]) ++bulls;
            else {
                if (m[secret[i]]++ < 0) ++cows;
                if (m[guess[i]]-- > 0) ++ cows;
            }
        }
        return to_string(bulls) + "A" + to_string(cows) + "B";
    }
};