LeetCode_10 Regular Expression Matching

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Similar question: None by now

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

We can use both recursive method dynamic programming to solve this problem. 
1.Recursive method:

public class Solution {
public boolean isMatch(String s, String p) {
if(p.length() == 0) return (s.length() == 0);
if(s.length() == 0){
if(p.length() > 1 && p.charAt(1) == '*') return isMatch(s, p.substring(2));
else return false;
}
if(p.length() > 1 && p.charAt(1) == '*'){
if(s.charAt(0) == p.charAt(0) || p.charAt(0) == '.') return isMatch(s.substring(1), p) || isMatch(s, p.substring(2));
return isMatch(s, p.substring(2));
}
if(s.charAt(0) == p.charAt(0) || p.charAt(0) == '.') return isMatch(s.substring(1), p.substring(1));
else return false;
}
}

2.Dynamic programming:

public class Solution {
public boolean isMatch(String s, String p) {
if(s == null || p == null) return false;
int m = s.length();
int n = p.length();
char[] sa = s.toCharArray();
char[] pa = p.toCharArray();
boolean[][] board = new boolean[m+1][n+1];
board[0][0] = true;
for(int jj = 2; jj <= n; jj++){
if(pa[jj-1] == '*'){
if(pa[jj-2] == '*') board[0][jj] = board[0][jj-1];
else board[0][jj] = board[0][jj-2];
}
}
for(int ii = 1; ii <= m; ii++){
for(int jj = 1; jj <= n; jj++){
if(pa[jj-1] == '*'){
if(jj == 1) continue;
boolean b1 = board[ii][jj-2];
boolean b2 = board[ii][jj-1];
boolean b3 = board[ii-1][jj];
board[ii][jj] = b1 || b2 || (b3 && (sa[ii-1] == pa[jj-2] || pa[jj-2] == '.'));
}
else{
if(sa[ii-1] == pa[jj-1] || pa[jj-1] == '.') board[ii][jj] = board[ii-1][jj-1];
}
}
}
return board[m]
;
}
}