Lettcode_299_Bulls and Cows

Lettcode_299_Bulls and Cows

You are playing the followingBulls and Cowsgame 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: 1bull and 3cows. (The bull is 8, the cows are 0, 1and 7.)

 

Write a function to return a hint according to the secret number and friend's guess, useAto indicate the bulls andBto 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 1in friend's guess is a bull, the 2nd or 3rd 1is 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）題意為給定兩串數字，其中的一串數字可以看作為密碼，另一串可看作為待猜測的數字，將猜對了的數字（位置和數字都相同）個數記為：個數＋A，將位置不對而包含在密碼中的數字個數記為：個數＋B。

（2）該題主要考察字元匹配問題。由於兩串數字中數的個數相同，所以，對於位置相同且數字亦相同的情況，只需遍曆一次就能得到bull的個數；而對於位置不同且含有相同數的情況，則需要進行特殊處理以得到cow的個數。一方面，可能某一個數在其中的某一串數中出現了多次；另一方面，位置相同且數相同的情況也需要進行過濾。此處，建立一個Map來進行儲存和過濾。其中，map的key為數串中的數，value為該數出現的次數。這樣通過一次遍曆即可將數和其個數儲存起來，然後再進行兩次遍曆，分別得到同一位置上數相同情況的個數和不同位置上數相同情況的個數，即為所求。詳情見下方代碼。

（3）希望本文對你有所協助。謝謝。

 

演算法代碼實現如下：

 

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";}}</character,></character,>