Lettcode_299_Bulls and Cows

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本文是在学习中的总结，欢迎转载但请注明出处：http://blog.csdn.net/pistolove/article/details/50768550

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"

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"

You may assume that the secret number and your friend‘s guess only contain digits, and their lengths are always equal

思路：

（1）题意为给定两串数字，其中的一串数字可以看作为密码，另一串可看作为待猜测的数字，将猜对了的数字（位置和数字都相同）个数记为：个数＋A，将位置不对而包含在密码中的数字个数记为：个数＋B。

（2）该题主要考察字符匹配问题。由于两串数字中数的个数相同，所以，对于位置相同且数字亦相同的情况，只需遍历一次就能得到bull的个数；而对于位置不同且含有相同数的情况，则需要进行特殊处理以得到cow的个数。一方面，可能某一个数在其中的某一串数中出现了多次；另一方面，位置相同且数相同的情况也需要进行过滤。此处，创建一个Map来进行存储和过滤。其中，map的key为数串中的数，value为该数出现的次数。这样通过一次遍历即可将数和其个数存储起来，然后再进行两次遍历，分别得到同一位置上数相同情况的个数和不同位置上数相同情况的个数，即为所求。详情见下方代码。

（3）希望本文对你有所帮助。谢谢。

算法代码实现如下：

import java.util.HashMap;
import java.util.Map;

public class Bulls_and_Cows {

	public static void main(String[] args) {
		System.err.println(getHint("1122", "1222"));
	}

	public static String getHint(String secret, String guess) {
		char[] se = secret.toCharArray();
		char[] gu = guess.toCharArray();

		int len = se.length;
		int bull = 0;
		int cow = 0;

		Map<Character, Integer> seHas = new HashMap<Character, Integer>();
		for (int i = 0; i < len; i++) {
			if (!seHas.containsKey(se[i])) {
				seHas.put(se[i], 1);
			} else {
				seHas.put(se[i], seHas.get(se[i]) + 1);
			}
		}

		Boolean[] b = new Boolean[len];
		for (int i = 0; i < len; i++) {
			if (se[i] == gu[i]) {
				b[i]  = true;
				seHas.put(gu[i], seHas.get(gu[i])-1);
				cow++;
			}else {
				b[i] = false;
			}
		}

		for (int j = 0; j < len; j++) {
			if(b[j] == false && seHas.get(gu[j])!=null && seHas.get(gu[j])>0) {
				bull ++;
				seHas.put(gu[j], seHas.get(gu[j])-1);
			}
		}
		
		return cow + "A" + bull + "B";
	}
}

Lettcode_299_Bulls and Cows

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原文地址：http://blog.csdn.net/pistolove/article/details/50768550