10. Regular Expression Matching

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

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        这个问题是一个典型的动态规划。我们设置一个数组dp，dp[i][j]在s[0...i) 符合正则表达式p[0..j)的情况下，为true，反之则为false。

        以下是状态转移方程：

当 p[j-1]!='*'

     dp[i][j]=i>0&&dp[i-1][j-1]&&(s[i-1]==p[j-1]||p[j-1]=='.')

否则

    (忽略.*或者x*)              (已经匹配了x)                    (匹配了x)

    dp[i][j]=dp[i][j-2]||(i>0&&dp[i-1][j]&&(s[i-1]==p[j-2]||p[j-2]=='.'))

class Solution {
public:
    bool isMatch(string s, string p) {
        int m=s.length();
        int n=p.length();
        vector<vector<bool>>dp(m+1,vector<bool>(n+1,false));
        dp[0][0]=true;
        for(int i=0;i<=m;i++){
            for(int j=1;j<=n;j++){
                if(p[j-1]!='*'){
                    dp[i][j]=i>0&&dp[i-1][j-1]&&(s[i-1]==p[j-1]||p[j-1]=='.');
                }
                else{
                    dp[i][j]=dp[i][j-2]||(i>0&&dp[i-1][j]&&(s[i-1]==p[j-2]||p[j-2]=='.'));
                }
            }
        }
        return dp[m][n];
    }
};