leetCode299.BullsandCows哈希-創(chuàng)新互聯(lián)

299. Bulls and Cows

創(chuàng)新互聯(lián)專注于云安企業(yè)網(wǎng)站建設(shè),響應(yīng)式網(wǎng)站開發(fā),商城網(wǎng)站建設(shè)。云安網(wǎng)站建設(shè)公司,為云安等地區(qū)提供建站服務(wù)。全流程定制設(shè)計,專業(yè)設(shè)計,全程項目跟蹤,創(chuàng)新互聯(lián)專業(yè)和態(tài)度為您提供的服務(wù)

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(jié):這里使用到了mutiset這一容器,該容器允許出現(xiàn)重復(fù)的值。

2016-08-13 13:21:15

另外有需要云服務(wù)器可以了解下創(chuàng)新互聯(lián)scvps.cn,海內(nèi)外云服務(wù)器15元起步,三天無理由+7*72小時售后在線,公司持有idc許可證,提供“云服務(wù)器、裸金屬服務(wù)器、高防服務(wù)器、香港服務(wù)器、美國服務(wù)器、虛擬主機(jī)、免備案服務(wù)器”等云主機(jī)租用服務(wù)以及企業(yè)上云的綜合解決方案,具有“安全穩(wěn)定、簡單易用、服務(wù)可用性高、性價比高”等特點與優(yōu)勢,專為企業(yè)上云打造定制,能夠滿足用戶豐富、多元化的應(yīng)用場景需求。

文章題目:leetCode299.BullsandCows哈希-創(chuàng)新互聯(lián)
本文地址:http://muchs.cn/article14/dhsjge.html

成都網(wǎng)站建設(shè)公司_創(chuàng)新互聯(lián),為您提供網(wǎng)站排名、小程序開發(fā)、商城網(wǎng)站網(wǎng)站導(dǎo)航、全網(wǎng)營銷推廣電子商務(wù)

廣告

聲明:本網(wǎng)站發(fā)布的內(nèi)容(圖片、視頻和文字)以用戶投稿、用戶轉(zhuǎn)載內(nèi)容為主,如果涉及侵權(quán)請盡快告知,我們將會在第一時間刪除。文章觀點不代表本網(wǎng)站立場,如需處理請聯(lián)系客服。電話:028-86922220;郵箱:631063699@qq.com。內(nèi)容未經(jīng)允許不得轉(zhuǎn)載,或轉(zhuǎn)載時需注明來源: 創(chuàng)新互聯(lián)

外貿(mào)網(wǎng)站建設(shè)