leetcode Wildcard Matching ,Regular Expression Matching 正则表达式匹配和通配符匹配

一 、通配符匹配(递归和非递归)：

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

class Solution {
public:
    bool isMatch(const char *s,const char *p) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
     /* recursive solution (TLE)
     if(*p == 0 && *s == 0){
            return true;
        }
        if(*s == 0){
            while(*p == '*') ++p;
            return !(*p);
        }
        if(*p == 0){
            return false;
        }
        int i = 0;
        for(; p[i] && s[i]; ++i){
            if(p[i] == s[i] || p[i] == '?')
                continue;
            if(p[i] == '*'){
                return isMatch(s+i,p+i+1) || isMatch(s+i+1,p+i);
            }
            else{
                return true;
            }
            
        }
        return isMatch(s+i,p+i);
        */
        //dp solution
        /*
        dp[i][j] = 1 means s[1~i] 和 p[1~j] match。
        so：
        if p[j] == '*' && (dp[i][j-1] || dp[i-1][j]) 
            dp[i][j] =  1 
        else p[j] = ? || p[j] == s[i]
            dp[i][j] = dp[i-1][j-1];
        else dp[i][j] = false;
        */
       int len_s = strlen(s);
       int len_p = strlen(p);
       const char* tmp = p;
       int cnt = 0;
       while(*tmp) if(*(tmp++) != '*') cnt++;
       if(cnt > len_s) return false;
       
       vector<vector<bool> > dp(len_s+1,vector<bool>(len_p+1,false));
       dp[0][0] = true;
       for(int i = 1; i <= len_p; ++i){
           if(dp[0][i-1] && p[i-1] == '*') dp[0][i] = true;
           for(int j = 1; j <= len_s; ++j){
               if(p[i-1] == '*'){
                   dp[j][i] = dp[j-1][i] || dp[j][i-1];
               }
               else if(p[i-1] == '?' || p[i-1] == s[j-1]){
                   dp[j][i] = dp[j-1][i-1];
               }
               else{
                   dp[j][i] = false;
               }
           }
       }
       return dp[len_s][len_p];
    }
};

二、正则表达式匹配(递归和非递归)

Implement regular expression matching with support for'.'and'*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

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

The function prototype should be:
bool isMatch(const char *s,"a*") → true
isMatch("aa",".*") → true
isMatch("ab",".*") → true
isMatch("aab","c*a*b") → true

class Solution {
public:
    bool isMatch(const char *s,const char *p) {
        // Note: The Solution object is instantiated only once and is reused by each test case.    
       /*recursive solution
       if(0 == *p) return 0 == *s;
        
        if(*(p+1) != '*'){
            if(*p == *s || (*p == '.' && 0 != (*s))){
                return isMatch(s+1,p+1);
            }
            return false;
        }
        else{
            while(*p == *s || (*p == '.' && 0 != (*s))){
                if(isMatch(s,p+2)){
                    return true;
                }
                s++;
            }
            return isMatch(s,p+2);
        }
        */
        
        //dp solution(sunbaigui)
        int len_s = strlen(s);
        int len_p = strlen(p);
        vector<vector<bool> > dp(len_s+1,false));
        dp[0][0] = true;
        for(int i = 0; i <= len_s; ++i){
            char preChar = '\0';
            int preIdx = 0;
            for(int j = 1; j <= len_p; ++j){
                if(0!=i && (p[j-1] == '.' || p[j-1] == s[i-1])){
                    dp[i][j] = dp[i-1][j-1];
                }
                else if(p[j-1] == '*'){
                    if(0 != i && (preChar == '.' || preChar == s[i-1])){
                        dp[i][j] = dp[i-1][j] || dp[i][j-1];
                    }
                    dp[i][j] = dp[i][j] || dp[i][preIdx];
                }
                if(p[j-1] != '*') {
                    preChar = p[j-1];
                    preIdx = j-1;
                }
            }
        }
        return dp[len_s][len_p];
        
    }
};