10.Regular Expression Matching (String; Back-Track,DP)

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, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

 思路：如果下个字符是'*'，那么可以重复当前字符0,1,2...次=>带回溯的递归。O(n)=kn，k是*出现的次数。

class Solution {
public:
    bool isMatch(string s, string p) {
        return dfsIsMatch(s,p,0,0);
    }
    
    bool dfsIsMatch(const string& s, const string& p, int sIndex, int pIndex){
        if (p[pIndex] == '\0') //结束条件：s若是'\0'，p必须也是'\0'
            return  s[sIndex] == '\0';
        
        if (p[pIndex+1] == '*') {
            /* '.' means any character (except '\0')
             * '.' means repeat 0 or more times
             * '.*' means repeat '.' 0 or more times
             */ 
            while ((s[sIndex] != '\0' && p[pIndex] == '.') || s[sIndex] == p[pIndex]) { //'.'可以与除'\0'以外的任何字符匹配
                if (dfsIsMatch(s, p, sIndex, pIndex+2)) //p[pIndex] repeat 0 times
                    return true;

                sIndex += 1;//p[pIndex]在原基础上repeat次数+1
            }
            
            return dfsIsMatch(s, p, sIndex, pIndex+2); //when s[sIndex] != p[pIndex] && p[pIndex] != '.'(此时只有s[sIndex]==p[pIndex]=='\0'时可能return true)
        } 
        else if ((s[sIndex] != '\0' && p[pIndex] == '.') || s[sIndex] == p[pIndex]) {
            return dfsIsMatch(s, p, sIndex + 1, pIndex + 1);
        }
       
        return false;
    }
};

 

法II：动态规划。设二位数组dp[i][j]来表示两个string的状态关系，dp[i][j]=true，表示s[0..i-1]匹配p[0..j-1]的结果。O(n)=n2

dp的下标0表示NULL，设置的原因是，比方如果p的第二个字符是*，那么p可以从第三个字符开始与s的第一个字符匹配，也就是dp[0][2]=true。

class Solution {
public:
    bool isMatch(string s, string p) {
        int sLen=s.length();
        int pLen = p.length();
        bool dp[sLen+1][pLen+1]={false}; //[0]for string NULL
        
        //initialize
        dp[0][0]=true;
        for(int j = 1; j <= pLen; j++){
            dp[0][j]=j>=2 && dp[0][j-2] && p[j-1]=='*'; //* each other letter
        }
        
        //state transfer
        for(int i = 1; i <= sLen; i++){
            for(int j = 1; j <= pLen; j++){
                if(p[j-1]=='*'){
                    dp[i][j]=(j>=2 && dp[i][j-2])/* repeat 0 times*/
                    ||(dp[i][j-1])/*repeat 1 time*/
                    ||(j>=2 && (s[i-1]==p[j-2] || p[j-2]=='.') && dp[i-1][j])/*repeat several times*/;
                }
                else{ //p[j-1]!='*'
                    dp[i][j]=dp[i-1][j-1] && (s[i-1]==p[j-1] || p[j-1]=='.');
                }
            }
        }
        
        return dp[sLen][pLen];
    }
};