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 preceding 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

 1 class Solution {
 2     public boolean isMatch(String s, String p) {
 3         int n = s.length(), m = p.length();
 4         boolean [][]f = new boolean[n + 2][m + 2];
 5         f[0][0] = true;
 6         for (int j = 1; j <= m; ++j) {
 7             for (int i = 0; i <= n; ++i) {
 8                 if (p.charAt(j - 1) >= 'a' && p.charAt(j - 1) <= 'z') {
 9                     if (i >= 1 && 
10                         p.charAt(j - 1) == s.charAt(i - 1) && f[i - 1][j - 1]) {
11                         f[i][j] = true;
12                     } 
13                 } else if (p.charAt(j - 1) == '.') {
14                     if (i >= 1 && f[i - 1][j - 1]) {
15                         f[i][j] = true;
16                     }
17                 } else if (p.charAt(j - 1) == '*') {
18                     if (f[i][j - 1]) f[i][j] = true;
19                     if (i >= 1 
20                         && (p.charAt(j - 2) == '.' 
21                             ||p.charAt(j - 2) == s.charAt(i - 1) )
22                         && f[i - 1][j - 1]) 
23                         f[i][j] = true;
24                     if (f[i][j - 2]) f[i][j] = true;
25                     if (i >= 1 && (p.charAt(j - 2) == '.' ||
26                         p.charAt(j - 2) == s.charAt(i - 1) )
27                         && f[i - 1][j]) 
28                         f[i][j] = true;
29                 }
30             }
31         }
32         return f[n][m];
33     }
34 }
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posted @ 2020-02-02 21:23  hyx1  阅读(154)  评论(0)  编辑  收藏  举报