文章--LeetCode算法--TwoSumRegularExpressionMatching

目录

• RegularExpressionMatching
  • 问题描述
  • 实例
  • 实现代码

RegularExpressionMatching

问题描述

Implement regular expression matching with support for '.' and '*'.

实例

'.' 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", "cab") ? true

实现代码

 
    public class Solution {
        public boolean isMatch(String str, String regex) {
            boolean[][] dp = new boolean[str.length() + 1][regex.length() + 1];
            dp[0][0] = true;
            for (int i = 1; i < regex.length() + 1; i++) {
                if (regex.charAt(i - 1) == '*')
                    dp[0][i] = dp[0][i - 2];
            }
            for (int i = 1; i < dp.length; i++) {
                for (int j = 1; j < dp[0].length; j++) {
                    if (match(str.charAt(i - 1), regex.charAt(j - 1))) {
                        dp[i][j] = dp[i - 1][j - 1];
                    }
                    else {
                        if (regex.charAt(j - 1) == '*') {
                            dp[i][j] = dp[i][j - 2];
                            if (match(str.charAt(i - 1), regex.charAt(j - 2))) {
                                dp[i][j] |= dp[i - 1][j];
                            }
                        }
                    }
                }
            }
            return dp[str.length()][regex.length()];
        }
        
        private boolean match(char c1, char r) {
            return c1 == r || r == '.';
        }
    }