44.wildcard matching

问题描述：

Given an input string (s) and a pattern (p), 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).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like ? or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

Input:
s = "acdcb"
p = "a*c?b"
Output: false

思路：

这道题一开始我以为可以像正则匹配那样解，然而我太天真，超时了。。所以就没有想法了。。

看了大佬的解法，但是还是不懂：）

然后又看了大佬2，大佬3的解释，感觉稍稍明白了一些。

当指针所指的字符不为*时非常好办：

　　1. *pcur == '?' : pcur++, scur++

　　2. *scur == *pcur ： pcur++, scur++

　　3. *scur != *pcur: 失配

当指针所指的为*时：

　　是将pstar后的字符串与sstar开始的各种可能的字符串进行匹配

　　即：p[j]="*"，尝试匹配p[j+1..p.length]和s[i..s.length]、s[i+1..s.length]、s[i+2..s.length]...

　　对最后一个*的后面的字符串进行匹配。

代码：

class Solution {
public:
    bool isMatch(string s, string p) {
        char *scur = &s[0], *pcur = &p[0], *sstar = NULL, *pstar = NULL;
        while(*scur){
            if(*scur == *pcur || *pcur == '?'){
                scur++;
                pcur++;
            }else if(*pcur == '*'){
                pstar = pcur++;
                sstar = scur;
            }else if(pstar){
                pcur = pstar + 1;
                scur = ++sstar;
            }else{
                return false;
            }
        }
        while(*pcur == '*')
            ++pcur;
        return !*pcur;
    }
};

动态规划：

emmmmmmmm……

现在不想做(⁎⁍̴̛ᴗ⁍̴̛⁎)