Leetcode：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

递归超时

bool isMatchR(const char *s, const char *p)
    {
        if(NULL==s && NULL==p)
            return true;
        if(NULL==s || NULL==p)
            return false;
        if(*s=='\0' && *p=='\0')
            return true;
        if(*s=='\0' || *p=='\0')
            return false;

        if(*p == '*')
        {
            while(*(p+1)=='*')
                ++p;    
            if(*(p+1)=='\0')
                return true;
            return isMatch(s,p+1)   || isMatch(s+1,p);
        }
        else if(*p == '?')
        {
            return isMatch(s+1,p+1);    
        }
        else
        {
            if(*p!=*s)  
                return false;
            return isMatch(s+1,p+1);
        }
    }

动态规划超内存

bool isMatchDP(const char *s, const char *p)
    {
        if(NULL==s && NULL==p)
            return true;
        if(NULL==s || NULL==p)
            return false;
        int lens = strlen(s),lenp = strlen(p);
        if(lens==0 && lenp==0)
            return true;
        else if(lens==0 || lenp==0)
            return false;
        bool** dp = new bool*[lens];
        for(int i=0;i=0 && j-1>=0)
                        dp[i][j] = dp[i-1][j-1];  
                    else
                        dp[i][j] = false;
                }
                else
                {
                    dp[i][j] = false;   
                }
            }
        }
        bool bcode = dp[lens-1][lenp-1];
        for(int i=0;i

贪心通过：

 bool isMatch(const char *s, const char *p)
    {
        const char* star=NULL;
        const char* ss=s;
        while(*s)
        {
            if(*p=='*')
            {
                star=p;++p;ss=s;    
            }
            else if(*p=='?' || *p==*s)
            {
                ++p;++s;
            }
            else if(NULL!=star)
            {
                p=star+1;s=++ss;    
            }
            else
                return false;
        }
        while(*p=='*')
            ++p;
        return !*p;
        
    }

这题做的很神伤啊，参考：

http://tech-wonderland.net/blog/leetcode-wildcard-matching.html

Wildcard Matching Solution and Precautions:

(1) recursive solution or DP, F(i, j) denote if s[0]…s[i-1] and p[0]…p[j-1] is matched or not

 F(i, j) = (s[i-1] == p[j-1]) && F(i-1, j-1) if  p[j-1] != ‘*’ or ‘?’
F(i, j) = F(i-1, j-1)  if  p[j] == ‘?’
F(i, j) = OR F(i-l, j-1) for l = 0, … i,  if  p[j] == ‘*’

(2) greedy: whenever encounter ‘*’ in p, keep record of the current position of ‘*’ in p and the current index in s. Try to match the stuff behind this ‘*’ in p with s, if not matched, just s++ and then try to match again.

http://blog.csdn.net/lifajun90/article/details/10582733