中文字幕av专区_日韩电影在线播放_精品国产精品久久一区免费式_av在线免费观看网站

溫馨提示×

溫馨提示×

您好,登錄后才能下訂單哦!

密碼登錄×
登錄注冊×
其他方式登錄
點擊 登錄注冊 即表示同意《億速云用戶服務條款》

leetCode 299. Bulls and Cows 哈希

發布時間:2020-07-14 16:30:02 來源:網絡 閱讀:506 作者:313119992 欄目:編程語言

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 01 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

向AI問一下細節

免責聲明:本站發布的內容(圖片、視頻和文字)以原創、轉載和分享為主,文章觀點不代表本網站立場,如果涉及侵權請聯系站長郵箱:is@yisu.com進行舉報,并提供相關證據,一經查實,將立刻刪除涉嫌侵權內容。

AI

广灵县| 准格尔旗| 鹤岗市| 仲巴县| 松滋市| 琼海市| 南部县| 来凤县| 杨浦区| 家居| 卢氏县| 亳州市| 宣化县| 大冶市| 凤城市| 成都市| 遵化市| 潢川县| 响水县| 南澳县| 巴马| 蓬莱市| 长宁县| 宁蒗| 永和县| 桓台县| 霍山县| 平昌县| 思茅市| 象州县| 彝良县| 开江县| 平江县| 北宁市| 乌苏市| 措勤县| 东乡县| 依安县| 香格里拉县| 彰武县| 上思县|