299. Bulls and Cows

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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.

思路：

1.遍历两个字符串A,B，如果某一位置的字符相等，那么bull++，否则将A中的字符放入一个vector中，B中的字符放入一个mutiset中。

2.遍历vector A，如果在B的mutiset中找到相等的元素，cow++，将这个元素从mutiset中删除。

3.组织字符串。

代码如下：

class Solution {
public:
    string getHint(string secret, string guess) 
    {
    	vector secretVector;
    	multiset guessMulSet;
    	int i;
    	int bulls = 0;
    	int cows = 0;
    	for (i = 0; i < secret.size(); i++)
    	{
    		if (secret[i] == guess[i])
    			bulls++;
    		else
    		{
    			secretVector.push_back(secret[i]);
    			guessMulSet.insert(guess[i]);
    		}
    	}
    	int vecLen = secretVector.size();
    	for (i = 0; i < vecLen; i++)
    	{
    		if (guessMulSet.find(secretVector[i]) != guessMulSet.end())
    		{
    			cows++;
    			guessMulSet.erase(guessMulSet.find(secretVector[i]));
    		}
    	}
    	string result = "";
    	stringstream ss;
    	stringstream ss1;
    	ss << bulls;
    	string bullStr = ss.str();
    
    	ss1 << cows;
    	string cowStr = ss1.str();
    	result += bullStr;
    	result += "A";
    	result += cowStr;
    	result += "B";
    
    	return result;
    }
};

总结：这里使用到了mutiset这一容器，该容器允许出现重复的值。

2016-08-13 13:21:15

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标题名称：leetCode299.BullsandCows哈希-创新互联
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