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Regular Expression Matching leetcode solution.

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 Regular Expression Matching leetcode solution.

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

'.' Matches any single character.​​​​
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).

 

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
 

Constraints:

1 <= s.length <= 20
1 <= p.length <= 30
s contains only lowercase English letters.
p contains only lowercase English letters, '.', and '*'.
It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.


class Solution {
int [][]hm;

public boolean isMatch(String s, String p) {
    hm=new int[s.length()+1][p.length()+1];
    return dfs(s,p,0,0);
}
boolean dfs(String s,String p,int i,int j)
{
    if(hm[i][j]!=0)
        return hm[i][j]==1;
   
    boolean ans;
    
    if(j>=p.length())
        ans= i==s.length();
    else
    {
        boolean match=(i <s.length() && (s.charAt(i)==p.charAt(j) || p.charAt(j)=='.'));

        if(j+1<p.length() && p.charAt(j+1)=='*')
            ans=dfs(s,p,i,j+2) || (match && dfs(s,p,i+1,j));
        else
            ans=match && dfs(s,p,i+1,j+1);
    }
    hm[i][j]=ans? 1 : 2;
    return ans;
    
}
}





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