10. Regular Expression Matching
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 <= 30
- 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.
class Solution {
public boolean isMatch(String s, String p) {
if (s == null || p == null) {
return false;
}
boolean[][] dp = new boolean[s.length()+1][p.length()+1];
dp[0][0] = true;
for(int i=0;i<p.length();i++){
if(p.charAt(i)=='*'&&dp[0][i-1]){
dp[0][i+1] = true;
}
}
for(int i=0;i<s.length();i++){
for(int j=0;j<p.length();j++){
if(p.charAt(j)=='.'){
dp[i+1][j+1] = dp[i][j];
}
if(p.charAt(j)==s.charAt(i)){
dp[i+1][j+1] = dp[i][j];
}
if(p.charAt(j)=='*'){
if(p.charAt(j-1)==s.charAt(i)||p.charAt(j-1)=='.'){
dp[i+1][j+1] = dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1];
}else{
dp[i+1][j+1] = dp[i+1][j-1];
}
}
}
}
return dp[s.length()][p.length()];
}
}