(java)Bulls and Cows

最新推荐文章于 2022-12-04 12:25:49 发布

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最新推荐文章于 2022-12-04 12:25:49 发布

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分类专栏： leetcode(java实现) 文章标签： leetcode java Bulls and Cows

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本文介绍了一个有趣的数字猜谜游戏——Bulls and Cows，并提供了一个实现该游戏逻辑的代码示例。玩家需要根据提示逐步猜出隐藏的数字。文章详细解释了如何通过比较猜测和秘密数字来计算提示信息，包括正确数字和位置的“牛”（bulls）以及正确数字但错误位置的“羊”（cows）。通过实例演示，读者可以深入理解并实现这一经典游戏。

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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:设置两个flag数组，flag1和flag2,，分别表示secret中的位置和guess中的位置是否被计数过。

先遍历一遍secret和guess早出相同位置上相同元素的个数。并将这些位的flag1和flag2标志为1,。然后再遍历secret和guess寻找不同位置上相同元素的个数，找到一对标志一对，防止重复计数。

代码如下(已通过leetcode)

public class Solution {
public String getHint(String secret, String guess) {
int countA=0,countB=0;
int[] flag1=new int[secret.length()];
int[] flag2=new int[secret.length()];
int i=0,j=0;
for(int k=0;k<secret.length();k++) {
if(secret.charAt(k)==guess.charAt(k)) {
countA++;
flag1[k]=1;
flag2[k]=1;
}
}
for(i=0;i<secret.length();i++) {
for(j=0;j<secret.length();j++) {
if(flag1[i]!=0) break;
else {
if(flag2[j]!=0) continue;
else {
if(secret.charAt(i)==guess.charAt(j)) {
if(i!=j) {
countB++;
flag1[i]=1;
flag2[j]=1;
}
}
}
}
}
}
return countA+"A"+countB+"B";
}
}