标签：string 情况 exp span als har aaa should examples
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
1.p为空，那s必然也为空
2.s为空，看p的第二字符是否为*，为*的话 则为isMatch(s,p.substring(2));
3.都不为空，p的第二字符为*，一种情况是直接跳过*部分isMatch(s,p.substring(2))，一种情况是首字符相等isMatch(s.substring(1),p)，一种情况是p首字符是.
4.都不为空，p的第二字符不为*，看首字符是否相等
1 public class Solution { 2 public boolean isMatch(String s, String p) { 3 if ("".equals(p)) 4 return "".equals(s); 5 else if ("".equals(s)){ 6 if (p.length()>1&&p.charAt(1)==‘*‘){ 7 return isMatch(s,p.substring(2)); 8 }else 9 return false; 10 } 11 else if (p.length()>1&&p.charAt(1)==‘*‘){ 12 if (isMatch(s,p.substring(2)))//skip .* 13 return true; 14 else if(s.charAt(0)==p.charAt(0)||p.charAt(0)==‘.‘){ 15 return isMatch(s.substring(1),p); 16 }else { 17 return false; 18 } 19 }else { 20 if (p.charAt(0)==s.charAt(0)||p.charAt(0)==‘.‘) 21 return isMatch(s.substring(1),p.substring(1)); 22 else 23 return false; 24 } 25 } 26 }
leetcode 10. Regular Expression Matching
标签：string 情况 exp span als har aaa should examples
原文地址：http://www.cnblogs.com/sure0328/p/7150605.html
