leetcode 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.
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.
这道题还是有些难度的。
首先我们得到bulls的数量还是非常容易的。只要比较s[i]==t[i]如果符合就加1就好了。
得到cows的数量比较难,刚开始我的想法是只要在s.find(t[i])!=-1就可以加1了。但是遇到
1234
0111
这种情况的时候,答案就错了,答案应该是0A1B而以上的方法得到确实0A3B
所以要对每一个元素的次数进行统计,例如s中的'1'已经被匹配过了,就不可以再被匹配了。
但是如果符合s[i]==t[i]的话,是属于bulls的。所以次数再减减。
最后一个注意的点是:int转string
最开始我用的方法是 char a=b+'0';但是这样的方法不能有大于9的数字出现,否则就会出错,因为‘0’+10就会变成':'而不是'10'
最后在网上查的方法:
#include<sstream> int a; string str; stringstream ss; ss<<a; ss>>str;