時(shí)間復(fù)雜度O(mn),dp[i][j] 代表字符串s的前i個(gè)字符和字符串p的前j個(gè)字符是否匹配，可以匹配兩個(gè)字符串均包含通配符的情況

class Solution {
public:
    /**
     * @param s: A string 
     * @param p: A string includes "?" and "*"
     * @return: A boolean
     */
    bool isMatch(const char *s, const char *p) {
    // write your code here
    
    if(s == NULL && p == NULL){
        return true;
    }
    
    if(s == NULL || p == NULL){
        return false;
    }
        
    
    int sl = (int)strlen(s);
    int pl = (int)strlen(p);
    //bool dp[][] = new bool[sl+1][pl+1];
    bool dp[sl+1][pl+1];
    for(int i = 1; i <= sl; ++i){
        for(int j = 1; j <= pl; ++j){
            dp[i][j] = false;
        }
    }
    dp[0][0] = true;
    for(int i = 1; i <= sl; ++i){
        if(dp[i-1][0] == true && s[i-1] == '*'){
            dp[i][0] = true;
        }else{
            dp[i][0] = false;
        }
    }
    
    for(int i = 1; i <= pl; ++i){
        if(dp[0][i-1] == true && p[i-1] == '*'){
            dp[0][i] = true;
        }else{
            dp[0][i] = false;
        }
    }
    
    for(int i = 1; i <= sl; ++i){
        for(int j = 1; j <= pl; ++j){
            if(s[i-1] == '*' || p[j-1] == '*'){
                dp[i][j] = dp[i-1][j] || dp[i][j-1];
            }else if(s[i-1] == '?' || p[j-1] == '?'){
                dp[i][j] = dp[i-1][j-1];
            }else {
                dp[i][j] = ((s[i-1] == p[j-1] ? true : false) && dp[i-1][j-1]);
            }
            
        }
    }
    
   // cout << dp[sl-1][pl-1] << endl;
    
    return dp[sl][pl];
}
};