带有通配符的字符串匹配算法－C/C++

//方法一，从无通配符到有?再到有*，逐步推进分析
char strMatch( const char *str1, const char *str2)  
{  
	int slen1 = strlen(str1);  
	int slen2 = strlen(str2);
	
	//实际使用时根据strl的长度来动态分配表的内存
	char matchmap[128][128];
	memset(matchmap, 0, 128*128); 
	matchmap[0][0] = 1;  
	int i, j, k;  
	//遍历目标字符串符串
	for(i = 1; i<= slen1; ++i)  
	{  
		//遍历通配符串
		for(j = 1; j<=slen2; ++j)
		{
			//当前字符之前的字符是否已经得到匹配
			if(matchmap[i-1][j-1])
			{
				//匹配当前字符
				if(str1[i-1] == str2[j-1] || str2[j-1] == '?')
				{ 
					matchmap[i][j] = 1; 
					//考虑星号在末尾的情况
					if( i  == slen1 && j < slen2)
					{
						for ( k = j+1 ; k <= slen2 ; ++k )
						{
							if( '*' == str2[k-1])
							{
								matchmap[i][k] = 1;
							}else{
								break;
							}
						}
					}
				}else if(str2[j-1] == '*')
				{
					//遇到星号，目标字符串到末尾都能得到匹配
					for(k = i-1; k<=slen1; ++k)
					{
						matchmap[k][j] = 1;  
					}
				}
			}
		}
		//如果当前字符得到了匹配则继续循环，否则匹配失败
		for(k = 1; k<=slen2; ++k)  
		{
			if(matchmap[i][k])
			{
				break; 
			}
		}
		if(k>slen2)
		{
			return 0;  
		}
	}
	return matchmap[slen1][slen2];  
}  
//方法二，分析每个情况。
char strMatch( const char *str1, const char *str2)  
{  
	int slen1 = strlen(str1);  
	int slen2 = strlen(str2);
	
	//实际使用时根据strl的长度来动态分配表的内存
	char matchmap[128][128];
	memset(matchmap, 0, 128*128);  
	int i, j, k;  
	//定义内循环的范围
	int lbound = 0;
	int upbound = 0;
	//遍历目标字符串符串
	for(i = 0; i< slen1; ++i)  
	{
		//遍历通配符串
		int bMatched = 0;
		int upthis = upbound;
		for(j = lbound; j<=upthis ; ++j)
		{
			//匹配当前字符
			if(str1[i] == str2[j] || str2[j] == '?')
			{ 
				matchmap[i][j] = 1;
				if(0 == bMatched)
				{
					lbound = j+1;
				}
				upbound = j+1;
				bMatched = 1;
				if(i == slen1 - 1)
				{
					//考虑末尾是*的特殊情况
					for(k = j+1 ; k < slen2 && '*' == str2[k] ; ++k)
					{
						matchmap[i][k] = 1;
					}
				}
			}else if(str2[j] == '*')
			{
				if(0 == bMatched)
				{
					lbound = j;
				}
				//遇到星号，目标字符串到末尾都能得到匹配
				for(k = i; k< slen1; ++k)
				{
					matchmap[k][j] = 1;  
				}
				k = j;
				while( '*' == str2[++k])
				{
					matchmap[i][k] = 1;
				}
				if(str1[i] == str2[k] || str2[k] == '?')
				{
					matchmap[i][k] = 1;
					upbound = k+1;
					if(i == slen1 - 1)
					{
						//考虑末尾是*的特殊情况
						for(k = k+1 ; k < slen2 && '*' == str2[k] ; ++k)
						{
							matchmap[i][k] = 1;
						}
					}
				}else{
					upbound = k;
				}
				bMatched = 1;
			}
		}
		//居然没有匹配到
		if(!bMatched )
		{
			return 0;
		}
	}
	return matchmap[slen1-1][slen2-1];  
}