[剑指offer]正则表达式匹配

 

题目描述

请实现一个函数用来匹配包括'.'和'*'的正则表达式。模式中的字符'.'表示任意一个字符，而'*'表示它前面的字符可以出现任意次（包含0次）。 在本题中，匹配是指字符串的所有字符匹配整个模式。例如，字符串"aaa"与模式"a.a"和"ab*ac*a"匹配，但是与"aa.a"和"ab*a"均不匹配

 

 

题目链接：

https://www.nowcoder.com/practice/45327ae22b7b413ea21df13ee7d6429c?tpId=13&tqId=11205&rp=3&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

 

 

 

 

 

package com.sunshine.OFFER66_SECOND;

import org.junit.Test;

public class A52_match {

    @Test
    public void test() {
        char[] str = {'a', 'a', 'a'};
        char[] pattern = {'a', 'b', '*', 'a'};
        boolean match = match(str, pattern);
        System.out.println(match);
    }

    public boolean match(char[] str, char[] pattern) {
        return judge(str, 0, pattern, 0);
    }

    public boolean judge(char[] str, int cur, char[] pattern, int pos) {
        if (pos == pattern.length && cur == str.length) {
            return true;
        }
        if (pos == pattern.length && cur < str.length) {
            return false;
        }
        if (pos < pattern.length && cur == str.length) {
            if (pos + 1 < pattern.length && pattern[pos + 1] == '*') {
                return judge(str, cur, pattern, pos + 2);
            }
            return false;
        }

        if (pattern[pos] == '.' || str[cur] == pattern[pos]) {
            if (pos + 1 < pattern.length && pattern[pos + 1] == '*') {
                return judge(str, cur + 1, pattern, pos) || judge(str, cur + 1, pattern, pos + 2) || judge(str, cur, pattern, pos + 2);
            } else {
                return judge(str, cur + 1, pattern, pos + 1);
            }
        }
        if (pattern[pos] != str[cur] && pos + 1 < pattern.length && pattern[pos + 1] == '*') {
            return judge(str, cur, pattern, pos + 2);
        }
        return false;
    }
}