通配符

？ 通配一个字符

*  通配零至多个字符

 

首先确定通配的字符串去掉*后，其长度比要匹配的字符串的长度小，这样就可以按照通配字符串来移动迭代器，而不必担心要匹配的字符串会越界。

第一个*以前的字符串要严格匹配，第一个*以后的字符串，要属于匹配字符串。

 

#include<iostream>
#include<string>
#include<list>
#include<algorithm>


using namespace std;

struct Op
{
	bool operator()(char a, char b)
	{
		if(b == '?')
		{
			return true;
		}
		else if( b == a)
		{
			return true;
		}
		else
		{
			return false;
		}
	}
};

int main()
{
	string youxing;
	string meixing;
	cin>>youxing>>meixing;
	
	//比较长度，避免后面担心匹配字符串越界
	list<char>ltemp;
	ltemp.assign(youxing.begin(),youxing.end());
	ltemp.remove('*');
	if(ltemp.size()>meixing.size())
	{
		cout<<"false"<<endl;
		return 0;
	}

	string::iterator iter1beg,iter1end,iter2beg,iter2end;
	iter1beg = youxing.begin();
	iter2beg = meixing.begin();
	iter1end = find(iter1beg,youxing.end(),'*');
	//第一个*之前的字符串严格匹配
	while(iter1beg!=iter1end)
	{
		if(*iter1beg != *iter2beg)
		{
			if(*iter1beg != '?')
			{
				cout<<"false"<<endl;
				return 0;
			}
		}
		++iter1beg;
		++iter2beg;
	}

	//第一个*以后的字符串要属于匹配字符串
	while(iter1end != youxing.end() && (iter1end+1) != youxing.end())
	{
		iter1beg = iter1end+1;
		iter1end = find(iter1beg,youxing.end(),'*');
		iter2beg = search(iter2beg,meixing.end(),iter1beg,iter1end,Op());
		if(iter2beg == meixing.end())
		{
			cout<<"false"<<endl;
			return 0;
		}
		else
		{
			iter2beg = iter2beg + (iter1end-iter1beg);
		}
	}
	if(iter1end == youxing.end())
	{
		if(iter2beg == meixing.end())
		{
			cout<<"true"<<endl;
		}
		else
		{
			cout<<"false"<<endl;
		}
	}
	else if((iter1end+1) == youxing.end())
	{
		cout<<"ture"<<endl;
	}

	return 0;
}

 

转载于:https://www.cnblogs.com/johnsblog/p/3763074.html