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Leetcode——299. Bulls and Cows
阅读量:4137 次
发布时间:2019-05-25

本文共 2085 字,大约阅读时间需要 6 分钟。

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
s2i(10,0); vector
g2i(10,0); string res; for(int i=0;i

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
m;//key represents the number,value represents the times of the number appearing unordered_map
mm; int bull = 0, cow = 0; string res = ""; for (int i = 0; i
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; }};

转载地址:http://hexvi.baihongyu.com/

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