含有通配符*的字符匹配（C语言）

含有通配符的字符匹配，采用贪心算法

//1	->	true
//0	->	false
int IsMatch(const char* reg, const char *str) {
	int r_len = (int)strlen(reg);
	int r_p = 0;
	int r_p_last = -1;
	int s_len = (int)strlen(str);
	int s_p = 0;
	int s_p_last = -1;

	while (s_p < s_len) {
		if (r_p < r_len && (*(str + s_p) == *(reg + r_p))) {
			r_p++;
			s_p++;
		}
		else if (r_p < r_len && (*(reg + r_p) == '*')) {
			r_p_last = r_p;
			r_p++;
			s_p_last = s_p;
		}
		else if (r_p_last > -1) {
			r_p = r_p_last + 1;
			s_p_last++;
			s_p = s_p_last;
		}
		else {
			return 0;//false
		}
	}
	while (r_p < r_len && (*(reg + r_p) == '*')) {
		r_p++;
	}
	if (r_p == r_len) {
		return 1;//true
	}
	return 0;//false
}

　　如果要匹配？的话，添加  

*(reg + r_p) =='?'

就行了

if (r_p < r_len && (*(str + s_p) == *(reg + r_p))) {

if (r_p < r_len && (*(str + s_p) == *(reg + r_p)||*(reg + r_p) =='?')) {

 注：之前匹配？的地方没写对，经过评论区的读者提醒，已经改正