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
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];
}
};