题目描述：
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
解答：
package com.jack.algorithm;
/**
* create by jack 2018/10/24
*
* @auther jack
* @date: 2018/10/24 20:30
* @Description: 规则表达式匹配
*/
public class RegularExpressionMatching {
/**
* 题目描述：
* https://leetcode.com/problems/regular-expression-matching/
*
* 代码解释：
* https://www.cnblogs.com/qiaozhoulin/p/4785077.html
* https://www.cnblogs.com/grandyang/p/4461713.html
* @param s
* @param p
* @return
*/
public static boolean isMatch(String s, String p) {
//创建二维数组，行数为：s.length() + 1，列数为：p.length() + 1
boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
//设置最后一行，最后一列的值为true
dp[s.length()][p.length()] = true;
//从最后的字符串往前遍历，这样可以解决*匹配0个或者多个字符的问题
for (int i = s.length(); i >= 0; i--){
//字符串的模式遍历
for (int j = p.length() - 1; j >= 0; j--){
//i小于s的长度，并且s的i位置的字符等于p的j位置的字符，或者p的j位置为“.”，则说明匹配
boolean first_match = (i < s.length() &&
(p.charAt(j) == s.charAt(i) ||
p.charAt(j) == '.'));
if (j + 1 < p.length() && p.charAt(j+1) == '*'){
dp[i][j] = dp[i][j+2] || first_match && dp[i+1][j];
} else {
dp[i][j] = first_match && dp[i+1][j+1];
}
}
}
return dp[0][0];
}
public static void main(String[] args) {
String s1 = "aa";
String s2 = "aab";
String s3 = "mississippi";
String s4 = "abcd";
String s5 = "mississippi";
String s6 = "ab";
String s7 = "aaa";
String s8 = "aaa";
String s9 = "aaa";
String s10 = "aaa";
String p1 = "a";
String p2 = "c*a*b";
String p3 = "mis*is*p*.";
String p4 = "d*";
String p5 = "mis*is*ip*.";
String p6 = ".*c";
String p7 = "aaaa";
String p8 = "a*a";
String p9 = "ab*ac*a";
String p10 = "ab*a";
//boolean flag = isMatch(s1, p1);
boolean flag = isMatch(s2, p2);
//boolean flag = isMatch(s3, p3);
//boolean flag = isMatch(s4, p4);
//boolean flag = isMatch(s5, p5);
//boolean flag = isMatch(s6, p6);
//boolean flag = isMatch(s7, p7);
//boolean flag = isMatch(s8, p8);
//boolean flag = isMatch(s9, p9);
//boolean flag = isMatch(s10, p10);
System.out.println("flag=" + flag);
}
}
源码：