leetcode刷题之旅（10） Regular Expression Matching

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本文介绍了一种实现正则表达式匹配的算法，重点讨论了如何处理特殊字符'.'和'*'，并通过多个示例解释了算法的工作原理。文章提供了两种不同的实现方式及其源代码。

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 *.

样例

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

思路分析

代码

public boolean isMatch(String s, String p) {

    if (s == null || p == null) {
        return false;
    }
    boolean[][] dp = new boolean[s.length()+1][p.length()+1];
    dp[0][0] = true;
    for (int i = 0; i < p.length(); i++) {
        if (p.charAt(i) == '*' && dp[0][i-1]) {
            dp[0][i+1] = true;
        }
    }
    for (int i = 0 ; i < s.length(); i++) {
        for (int j = 0; j < p.length(); j++) {
            if (p.charAt(j) == '.') {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == s.charAt(i)) {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == '*') {
                if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {
                    dp[i+1][j+1] = dp[i+1][j-1];
                } else {
                    dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);
                }
            }
        }
    }
    return dp[s.length()][p.length()];
}

方法二

public boolean isMatch(String s, String p) {
		if (p.contains(".") || p.contains("*")) {
			if (p.length() == 1 || p.charAt(1) != '*')
				return comp(s, p, s.length(), 0) && isMatch(s.substring(1), p.substring(1));
			for (int i = 0; i == 0 || comp(s, p, s.length(), i - 1); i++) {
				if (isMatch(s.substring(i), p.substring(2)))
					return true;
			}
		}
		return s.equals(p);
	}

	private boolean comp(String s, String p, int sLen, int i) {
		return sLen > i && (p.charAt(0) == s.charAt(i) || p.charAt(0) == '.');
	}