LeetCode: Wildcard Matching

这题第一次做用dfs，结果过不了large，网上寻找到答案，这里s和p指针是会变的，重点处理ptr遇到*的时候，这个时候s和p都往后挪，p重新定位，看s后面和p后面的是不是match，如果不match则s向后移，再继续看后面的是不是match，直到s到底为止，这里str和ptr是浮动指针，s和p相对静止，记录前一个邵点。

class Solution {
public:
    bool isMatch(const char *s, const char *p) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        bool star = false;
        const char *str, *ptr;
        for (str = s, ptr = p; *str != '\0'; str++, ptr++) {
            if (*ptr == '*') {
                star = true;
                s = str, p = ptr;
                while (*p == '*') p++;
                if (*p == '\0') return true;
                str = s-1, ptr = p-1;
            }
            else if (*ptr == '?') continue;
            else {
                if (*str != *ptr) {
                    if (!star) return false;
                    s++;
                    str = s-1, ptr = p-1;
                }
            }
        }
        while (*ptr == '*') ptr++;
        return *ptr == '\0';
    }
};

 while的可能更好懂

class Solution {
public:
    bool isMatch(const char *s, const char *p) {
        const char *s1, *p1;
        s1 = s, p1 = p;
        bool star = false;
        while (*s1 != '\0') {
            if (*s1 == *p1 || *p1 == '?') {
                s1++, p1++;
            }
            else {
                if (*p1 == '*') {
                    star = true;
                    p = p1, s = s1;
                    while (*p == '*') p++;
                    if (*p == '\0') return true;
                    p1 = p;
                }
                else {
                    if (!star) return false;
                    s++;
                    s1 = s, p1 = p;
                }
            }
        }
        while (*p1 == '*') p1++;
        return *p1 == '\0';
    }
};