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  • LeetCode 299. Bulls and Cows

    原题链接在这里:https://leetcode.com/problems/bulls-and-cows/

    题目:

    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 01 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.

    题解:

    猜字游戏 secret.charAt(i) == guess.charAt(i)时bull++.

    用map来记录character 和出现的次数. secret扫过数字,mark对应count加一,guess扫过数字,mark对应count减一.

    若是mark[secret.charAt(i) - '0'] 小于0, 说明guess 已经更新过这个位置,此时numCow++. 反过来若是 mark[guess.charAt(i) - '0'] 大于0, 说明secret 已经跟新过这个位置, 此时numCow++.

    Time Complexity: O(n). 只需要扫一遍两个string. Space: O(1). 一个长度为10的array.

    AC Java:

     1 public class Solution {
     2     public String getHint(String secret, String guess) {
     3         int numBull = 0;
     4         int numCow = 0;
     5         int [] mark = new int[10];
     6         for(int i = 0; i<secret.length(); i++){
     7             if(secret.charAt(i) == guess.charAt(i)){
     8                 numBull++;
     9             }else{
    10                 if(mark[secret.charAt(i) - '0']++ < 0){
    11                     numCow++;
    12                 }
    13                 if(mark[guess.charAt(i) - '0']-- > 0){
    14                     numCow++;
    15                 }
    16             }
    17         }
    18         String res = String.valueOf(numBull) + "A" + String.valueOf(numCow) + "B";
    19         return res;
    20     }
    21 }
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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/4929636.html
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