2019年4月3日华为实习生笔试第1题-交替输出字符串

以你之姓@ 2022-02-24 11:24 196阅读 0赞

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5NDc2NzU2NzY_size_16_color_FFFFFF_t_70]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5NDc2NzU2NzY_size_16_color_FFFFFF_t_70 1]

笔试的时候我只调出来20%，后面找到原因了，也是一些小问题，要气死了  
第二题也没有先做出来，还是该先做出来第一题再做第二题的

#include <iostream>
    #include <vector>
    #include <string>
    #include <list>
    using namespace std;
    /* 3 2,5,6,7,9,5,7 1,7,4,3,4 2,5,6,1,7,4,7,9,5,3,4,7 */
    
    //变量用完记得清零！！！！！
    int main()
    { 
    	int N;
    	cin >> N;
    	string str;
    	vector<string> vecStr;
    	while (cin >> str)
    	{ 
    		vecStr.push_back(str+',');
    	}
    
    	int MaxLen = 0;
    	for (int i = 0; i < vecStr.size(); i++)
    	{ 
    		if (vecStr[i].size() > MaxLen)
    		{ 
    			MaxLen = vecStr[i].size();
    		}
    	}
    
    	if (MaxLen % (2 * N) == 0)
    	{ 
    		MaxLen = MaxLen / (2 * N);
    	}
    	else
    		MaxLen = MaxLen / (2 * N) + 1;
    
    	string outStr;
    	vector<int> count(vecStr.size(),0);
    	vector<int> flag(vecStr.size(), 0);
    	for (int j = 0; j < MaxLen; j++)
    	{ 
    		for (int i = 0; i < vecStr.size(); i++)
    		{ 
    			string newStr = vecStr[i];
    			if (newStr.substr(count[i] * (2 * N)).size() >= N * 2)
    			{ 
    				outStr = outStr + newStr.substr(count[i] * (2 * N), N * 2);
    				count[i]++;
    			}
    			else if(flag[i] == 0)
    			{ 
    				outStr = outStr + newStr.substr(count[i] * (2 * N));
    				flag[i] ++;
    			}
    		}
    	}
    
    	cout << outStr.substr(0,outStr.size()-1) << endl;
    
    	return 0;
    }

按Ctrl+Z，结束输入  
![在这里插入图片描述][20190405191113210.png]

意思代码经博友的提醒发现有问题，只能对个位数的数字有效。  
在牛客网上找了一下结题思路，如下：  
查找逗号的出现的次数  
//附上别人的代码，我加了些注释

#include <iostream>
    #include <vector>
    #include <stdio.h>
    #include <string>
    using namespace std;
    
    int main()
    { 
    	int len;
    	cin >> len;
    	vector<string> allstr;
    	string strtemp;
    	int i, j, temp;
    	while (cin >> strtemp)
    	{  
    		allstr.push_back(strtemp);
    	}
    	long long sumlen = 0;
    	for (i = 0; i < allstr.size(); i++)
    	{  
    		sumlen += allstr[i].length();
    	}
    	string result;
    	int num = 0;
    	int l = 0;
    	while (sumlen)
    	{  
    		for (i = 0; i < allstr.size(); i++)
    		{  
    			if (allstr[i].length() != 0)
    			{  
    				l = 0;//统计需要截取的字符长度
    				num = 0;//统计，号次数
    				for (j = 0; j < allstr[i].length(); j++)
    				{  
    					l++;
    					if (allstr[i][j] == ',')
    						num++;//统计，号次数
    					if (num == len)
    						break; 
    				} 
    
    				if (num == len)//说明字符长度足够截取
    				{ 
    					result += allstr[i].substr(0, l);
    					allstr[i] = allstr[i].erase(0, l);
    				} 
    				else//如果截取到尾部添加，
    				{  
    					result += allstr[i];
    					result += ',';
    					allstr[i].erase(0);
    				}
    			}
    		}
    		sumlen = 0;
    		//重新获取所有字符串的长度
    		for (i = 0; i < allstr.size(); i++)
    		{  
    			sumlen += allstr[i].length();
    		}
    	}
    	result.erase(result.length() - 1);
    	cout << result << endl;     
    	return 0;
    }

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5NDc2NzU2NzY_size_16_color_FFFFFF_t_70]: /images/20220224/621f8327240f4bc5b0b772c4470ff656.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3p5NDc2NzU2NzY_size_16_color_FFFFFF_t_70 1]: /images/20220224/c6855b9d6b224f13bcd749a7b06b4e71.png
[20190405191113210.png]: /images/20220224/8f08b8d02c0e4e86a1323102374e8f13.png