Wildcard Matching

Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false 
分析:一个超时版本如下:
class Solution {
public:
    bool isMatch(const char *s, const char *p) {
        if(*s == '\0') return *p == '\0';
        if(*p == '\0') return *s == '\0';
        int i = 0, j = 0;
        for(; *(s+i) != '\0' && *(p+j) != '\0';){
            if(*(s+i) == *(p+j) || *(p+j) == '?'){
                i++;
                j++;
            }else if(*(p+j) == '*'){
                while(*(s+i) != '\0'){
                    if(isMatch(s+i, p+j+1))
                        return true;
                    i++;
                }
                return false;
            }else return false;
        }
        return *(s+i) == *(p+j);
    }
};

 迭代版本相比递归版本要快,但难度要大。这里参考leetcode.pdf的解答,用两套指针(str,ptr)和(s,p)。(str,ptr)用于探索,(s,p)用于保存探索前的状态,一遍当探索失败后恢复到原来状态。同时要注意一些细节的处理,比如多个连续的'*'。

class Solution {
public:
    bool isMatch(const char *s, const char *p) {
        bool star = false;
        const char *str, *ptr;
        for(str = s, ptr = p; *str != '\0'; str++, ptr++){
            switch(*ptr){
                case '?':
                    break;
                case '*':
                    star = true;
                    s = str, p = ptr;
                    while(*p == '*')p++;
                    if(*p == '\0') return true;
                    str = s-1;
                    ptr = p-1;
                    break;
                default:
                    if(*str != *ptr){
                        if(!star) return false;
                        s++;
                        str = s-1;
                        ptr = p-1;
                    }
            }
        }
        while(*ptr == '*')ptr++;
        return *ptr == '\0';
    }
};

 

posted on 2014-10-22 22:38  Ryan-Xing  阅读(134)  评论(0)    收藏  举报