一道c++小编程题，

雨点打透心脏的1/2处 2022-12-26 08:23 137阅读 0赞

题目:

编写一个小程序,从标准输入读入一系列string对象,寻找连续重复出现的单词,程序应该找出满足以下条件的单词的输入位置:该单词的后面紧跟着再次出现自己本身,跟 踪重复次数量多的单词及其重复次数.输出重复次数的最大值,  
例如.如果输入是:  
how now now now brown cow cow

则输出表明now这个单词出现了三次

本人解法：

#include <iostream>
    #include <string>
    #include <cctype>
    #include <vector>
    using namespace std;
    int main() { 	string s;
    	vector<string> vec;
    	int maxRepeat = 0;
    	while (cin >> s){
    		if(ispunct(s[s.size()-1])) s = s.substr(0,s.size() - 1);  //去除最后的标点符号
    		vec.push_back(s);
    	}
    	vector<string>::size_type i = 0;
    	vector<int> vecInt(vec.size(),1);
    	while (i != vec.size() - 1){
    		int j = i;
    		string str = vec[i];
    		while(i != vec.size() - 1)
    		{
    			if(str == vec[++i]) vecInt[j] ++;
    			else break;
    		}
    	}
    	vector<int>::size_type k = 0;
    	int max = vecInt[k];
    	int flag = 0;
    	while (k != vecInt.size() - 1){
    		if(max < vecInt[++k]){
    			max = vecInt[k];
    			flag = k;
    		}
    	}
    	cout << "The word of " <<  vec[flag] << " repeats: " << vecInt[flag] << " times." << endl;
    	cout << endl; 	return 0;
    
    } 自己的解法答案是对的 但从空间复杂度来说要复杂的多，所以在网上搜了下，寻找到另外一种解决办法，修改后的代码如下: #include <iostream>
    #include <string>
    #include <cctype>
    using namespace std;
    int main() { 	string nowWord,beforeWord,result;
    	cin >> nowWord;
    	if(ispunct(nowWord[nowWord.size() - 1]))nowWord = nowWord.substr(0,nowWord.size() - 1);
    	result = beforeWord = nowWord;
    	int count = 1,maxCount = 1;
    	while(cin >> nowWord){
    		if(ispunct(nowWord[nowWord.size() - 1]))nowWord = nowWord.substr(0,nowWord.size() - 1);
    		if (beforeWord == nowWord){
    			count++;
    		}
    		else{
    			beforeWord = nowWord;
    			count = 1;
    		}
    		if(count > maxCount){
    			maxCount = count;
    			result = nowWord;
    		}
    	}
    	if(maxCount == 1) cout << "There is no word repeating." << endl;
    	else cout << "The word of " << result << " repeats: " << maxCount << " times." << endl;
    	return 0;
    
    } 特在此分享给大家 参考或给予好的想法参考改进。。谢谢~~