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10. Regular Expression Matching - 正则表达式匹配

Tags - 题目标签

Description - 题目描述

EN:

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

  • '.' Matches any single character.​​​​
  • '*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

 

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 (.)".

 

Constraints:

  • 1 <= s.length <= 20
  • 1 <= p.length <= 20
  • s contains only lowercase English letters.
  • p contains only lowercase English letters, '.', and '*'.
  • It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

ZH-CN:

给你一个字符串 s 和一个字符规律 p，请你来实现一个支持 '.' 和 '*' 的正则表达式匹配。

  • '.' 匹配任意单个字符
  • '*' 匹配零个或多个前面的那一个元素

所谓匹配，是要涵盖 整个 字符串 s的，而不是部分字符串。

 

示例 1：

输入：s = "aa", p = "a"
输出：false
解释："a" 无法匹配 "aa" 整个字符串。

示例 2:

输入：s = "aa", p = "a*"
输出：true
解释：因为 '*' 代表可以匹配零个或多个前面的那一个元素, 在这里前面的元素就是 'a'。因此，字符串 "aa" 可被视为 'a' 重复了一次。

示例 3：

输入：s = "ab", p = ".*"
输出：true
解释：".*" 表示可匹配零个或多个（'*'）任意字符（'.'）。

 

提示：

  • 1 <= s.length <= 20
  • 1 <= p.length <= 20
  • s 只包含从 a-z 的小写字母。
  • p 只包含从 a-z 的小写字母，以及字符 . 和 *。
  • 保证每次出现字符 * 时，前面都匹配到有效的字符

Link - 题目链接

LeetCode - LeetCode-CN

Latest Accepted Submissions - 最近一次 AC 的提交

Language Runtime Memory Submission Time
typescript 72 ms 46.3 MB 2022/05/01 22:37 
function isMatch(s: string, p: string): boolean {
  const sLen = s.length;
  const pLen = p.length;

  const dp: boolean[][] = new Array(sLen + 1).fill(false).map(i => new Array(pLen + 1).fill(false));

  dp[0][0] = true;

  for (let j = 1; j < pLen + 1; j++) {
    dp[0][j] = p[j - 1] === '*' && dp[0][j - 2];
  }

  for (let i = 1; i < sLen + 1; i++) {
    for (let j = 0; j < pLen + 1; j++) {
      if (p[j - 1] === '*') {
        if (s[i - 1] === p[j - 2] || p[j - 2] === '.') {
          dp[i][j] = dp[i][j - 2] || dp[i - 1][j];
        } else {
          dp[i][j] = dp[i][j - 2];
        }
      } else {
        if (p[j - 1] === '.' || s[i - 1] === p[j - 1]) {
          dp[i][j] = dp[i - 1][j - 1];
        } else {
          // dp[i][j] = false;
        }
      }
    }
  }

  return dp[sLen][pLen];
};

My Notes - 我的笔记

与 剑指 Offer 19: 正则表达式匹配 一样。

参考Krahets 的题解：

function isMatch(s: string, p: string): boolean {
  const sLen = s.length;
  const pLen = p.length;

  const dp: boolean[][] = new Array(sLen + 1).fill(false).map(i => new Array(pLen + 1).fill(false));

  dp[0][0] = true;

  for (let j = 1; j < pLen + 1; j++) {
    dp[0][j] = p[j - 1] === '*' && dp[0][j - 2];
  }

  for (let i = 1; i < sLen + 1; i++) {
    for (let j = 0; j < pLen + 1; j++) {
      if (p[j - 1] === '*') {
        if (s[i - 1] === p[j - 2] || p[j - 2] === '.') {
          dp[i][j] = dp[i][j - 2] || dp[i - 1][j];
        } else {
          dp[i][j] = dp[i][j - 2];
        }
      } else {
        if (p[j - 1] === '.' || s[i - 1] === p[j - 1]) {
          dp[i][j] = dp[i - 1][j - 1];
        } else {
          dp[i][j] = false;
        }
      }
    }
  }

  return dp[sLen][pLen];
};