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  • leetcode 299. Bulls and Cows

    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.

    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. 

    Please note that both secret number and friend's guess may contain duplicate digits.

    Example 1:

    Input: secret = "1807", guess = "7810"
    
    Output: "1A3B"
    
    Explanation: 1 bull and 3 cows. The bull is 8, the cows are 0, 1 and 7.

    Example 2:

    Input: secret = "1123", guess = "0111"
    
    Output: "1A1B"
    
    Explanation: The 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow.

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

    题目难度:简单题

    题目思路:因为数字在0-9,所以可以直接建立哈希表(长度为10的数组,键为数字,值为字符串中这个数字的个数。初始化为0(个))。如果同一位置的数字相同,那么bull++, 否则将它们分别记录在两个哈希表中。最后,统计两个哈希表同键对应的值最小的那个,累加起来即为cow。

    C++代码如下:

     1 class Solution {
     2 public:
     3     string getHint(string secret, string guess) {
     4         int len1 = secret.length(), len2 = guess.length();
     5         int bull = 0, cow = 0;
     6         vector<int> sCnt(10, 0), gCnt(10, 0);
     7         if ((len1 != len2) || (len1 == 0)) return "0A0B";
     8         for (int i = 0; i < len1; ++i) {
     9             if (secret[i] == guess[i]) {
    10                 ++bull;
    11             } else {
    12                 ++sCnt[secret[i] - '0'];
    13                 ++gCnt[guess[i] - '0'];
    14             }
    15         }
    16         for (int i = 0; i < 10; ++i) {
    17             cow += min(sCnt[i], gCnt[i]);
    18         }
    19         return std::to_string(bull) + 'A' + std::to_string(cow) + 'B';
    20     }
    21 };
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  • 原文地址:https://www.cnblogs.com/qinduanyinghua/p/13656159.html
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