Leetcode——299. Bulls and Cows

2024-02-02 01:38

文章标签 leetcode 299 cows bulls

本文主要是介绍Leetcode——299. Bulls and Cows，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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 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.

解答

两个解法，先说一开始没想到的很简答的一个解法：

class Solution {
public:string getHint(string secret, string guess) {int bull=0,cow=0;vector<int> s2i(10,0);vector<int> g2i(10,0);string res;for(int i=0;i<secret.size();i++){if(guess[i]==secret[i])bull++;else{s2i[secret[i]-'0']++;g2i[guess[i]-'0']++;}}for(int i=0;i<10;i++)//只需统计s2i和g2i中每个i对应的最小的那个值cow+=min(s2i[i],g2i[i]);res=to_string(bull)+"A"+to_string(cow)+"B";return res;}
};

自己写的复杂的
把问题解法想复杂了！

class Solution1 {
public:string getHint(string secret, string guess) {int len = secret.length();unordered_map<int, int> m;//key represents the number，value represents the times of the number appearingunordered_map<int, int> mm;int bull = 0, cow = 0;string res = "";for (int i = 0; i<len; i++){if (secret[i] == guess[i])//cow numberbull++;else{m[secret[i] - '0']++;mm[guess[i] - '0']++;}}for (int i = 0; i<len; i++){if (m.find(guess[i] - '0') != m.end())//m has the number{if (mm[guess[i] - '0']>0&&m[guess[i] - '0']>0){cow++;mm[guess[i] - '0']--;m[guess[i] - '0']--;}}}res = to_string(bull) + "A" + to_string(cow) + "B";return res;}
};