Regular Expression Matching

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对于这种几乎最高难度的题，甚于word ladder II，要放平心态。

坚持，加油！

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

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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

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