leetcode 10: Regular Expression Matching

Regular Expression Matching

Total Accepted: 39732 Total Submissions: 192279

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]
字符串match, 因为涉及到*可以match 0到多个字符, 必然有backtracking. 再backtracking的过程中, 如果重复, 可以pruning, memo, dynamic programming 进行优化.

[CODE]

1. backtracking

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