leetcode-10 Regular Expression Matching

Given an input string (s) and a pattern (p), 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).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.

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 precedeng 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 (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

根据题意分析就是自己实现正则判断，对于输入的字符串s和正则字符串p进行校验：

(.*代表所有任意字符串肯定为true)

首先判断第一个字符是否相等 boolean firstMatch = (s.charAt(0) == p.charAt(0) )||(p.charAt(0) =='.')

题意中s只能为小写字母 p只能为小写字母或是 . （表示任意字母）或 *（表示0或n个前置值）

然后对s的第二个字符串进行判断

其中如果p的第二个字符串为* 则第一个字符串匹配且递归判断 s.substring(1) 和正则 p 或递归判断 s与正则 p.substring(2)

当p的第二个字母非*时递归判断 s.substring(1)和p.substring(1)

注意判断各种特殊情况即可完成

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

最后结果有点惨呀。。。。

leetcode-10 Regular Expression Matching

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