小旭讲解 LeetCode 10. Regular Expression Matching

Problem

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 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 (.)".

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

题目

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。

Video Explaination

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Explaination

能否用动态规划的方式求解？

确认问题能否用动态规划的方式求解——即分析问题能否划分成若干个重叠的子问题，并且能够层层递进的求解（最优子结构）。

其中要体现两个动态规划问题的性质：

• 重叠子问题性质
• 最优子结构性质

分类（按照状态空间递进的方式，仅仅提及两种）

• 线性DP
  • 状态空间以线性的方式进行递推求解
• 区间DP
  • 状态空间以“区间长度”的方式进行递推求解（可参考题目 Stone Game II）

本题属于线性DP问题。

Code

Dynamic Programming（线性DP）

class Solution {
public:
    bool isMatch(string s, string p) {
        vector<vector<int>> dp(s.size() + 1, vector<int>(p.size() + 1, 0));
        dp[s.size()][p.size()] = 1;
        for (int i = s.size(); i >= 0; --i) {
            for (int j = p.size() - 1; j >= 0; --j) {
                bool first_match = i < s.size() && (s[i] == p[j] || p[j] == '.');
                if (j < p.size() - 1 && p[j + 1] == '*') {
                    dp[i][j] = (first_match && (dp[i + 1][j] || dp[i][j + 2])) || (!first_match && (dp[i][j + 2]));
                } else {
                    dp[i][j] = first_match && dp[i + 1][j + 1];
                }
            }
        }
        return dp[0][0];
    }
};

Comlexity Analysis

Time Complexity：O(M * N) M is the length of string s, N is the length of string p.
Space Complexity: O(M * N)

Recursive

class Solution {
public:
    bool isMatch(string s, string p) {
        if (p.empty()) return s.empty();
        bool first_match = !s.empty() && (s[0] == p[0] || p[0] == '.');
        if (p.length() >= 2 && p[1] == '*') {
            return isMatch(s, p.substr(2)) || (first_match && isMatch(s.substr(1), p));
        } else {
            return first_match && isMatch(s.substr(1), p.substr(1));
        }
    }
};

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Problem：https://leetcode-cn.com/problems/regular-expression-matching

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