LeetCode 299 Bulls and Cows

Problem:

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.

For example:

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.

Summary:

这道题题意略难懂。。搞了半天才明白。。

题目要求输出secret string和friend's guess两个字符串中相同位数和不同位数，其中：

相同位数bull：代表两个字符串中字符和下标都相同的位数

不同位数cow：代表两个字符串中字符相同但下标不同的位数

Solution:

Hash Table分别记录两个字符串中相应位数不同时字符出现个数。

 class Solution {
 public:
     string getHint(string secret, string guess) {
         int len = secret.size();
         int sec[] = {}, gue[] = {};
         int bull = , cow = ;

         for (int i = ; i < len; i++) {
             int s = secret[i] - '', g = guess[i] - '';
             if (s == g) {
                 bull++;
             }
             else {
                 if (sec[g]) {
                     cow++;
                     sec[g]--;
                 }
                 else {
                     gue[g]++;
                 }

                 if (gue[s]) {
                     cow++;
                     gue[s]--;
                 }
                 else {
                     sec[s]++;
                 }
             }
         }

         return (to_string(bull) + "A" + to_string(cow) + "B");
     }
 };