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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.
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.
思路:
1.遍歷兩個字符串A,B,如果某一位置的字符相等,那么bull++,否則將A中的字符放入一個vector中,B中的字符放入一個mutiset中。
2.遍歷vector A,如果在B的mutiset中找到相等的元素,cow++,將這個元素從mutiset中刪除。
3.組織字符串。
代碼如下:
class Solution { public: string getHint(string secret, string guess) { vector<char> secretVector; multiset<char> guessMulSet; int i; int bulls = 0; int cows = 0; for (i = 0; i < secret.size(); i++) { if (secret[i] == guess[i]) bulls++; else { secretVector.push_back(secret[i]); guessMulSet.insert(guess[i]); } } int vecLen = secretVector.size(); for (i = 0; i < vecLen; i++) { if (guessMulSet.find(secretVector[i]) != guessMulSet.end()) { cows++; guessMulSet.erase(guessMulSet.find(secretVector[i])); } } string result = ""; stringstream ss; stringstream ss1; ss << bulls; string bullStr = ss.str(); ss1 << cows; string cowStr = ss1.str(); result += bullStr; result += "A"; result += cowStr; result += "B"; return result; } };
總結:這里使用到了mutiset這一容器,該容器允許出現重復的值。
2016-08-13 13:21:15
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