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
class Solution {
public boolean isMatch(String s, String p) {
boolean dp[][] = new boolean[s.length() + 1][p.length() + 1];
for (int i = 0; i <= s.length(); i++) {
for (int j = 0; j <= p.length(); j++) {
// Base cases start
if (i == 0 && j == 0)
dp[i][j] = true;
else if (j == 0)
dp[i][j] = false;
else if (i == 0) {
if (p.charAt(j - 1) != '*')
dp[i][j] = false;
else
dp[i][j] = j >= 2 ? dp[i][j - 2] : false;
} // Base cases end
else if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.')
dp[i][j] = dp[i - 1][j - 1];
else if (p.charAt(j - 1) == '*') {
dp[i][j] = dp[i][j - 2];
if (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.')
dp[i][j] = dp[i][j] || dp[i - 1][j];
} else
dp[i][j] = false;
}
}
return dp[s.length()][p.length()];
}
}