10. Regular Expression Matching(Leetcode每日一题-2020.06.20)

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Problem

Given an input string (s) and a pattern §, 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 *.

Example1

Input:
s = “aa”
p = “a”
Output: false
Explanation: “a” does not match the entire string “aa”.

Example2

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”.

Example3

Input:
s = “ab”
p = “."
Output: true
Explanation: ".” means “zero or more (*) of any character (.)”.

Example4

Input:
s = “aab”
p = “cab”
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches “aab”.

Example5

Input:
s = “mississippi”
p = “misisp*.”
Output: false

Solution

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