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
Solution:
class Solution { public boolean isMatch(String s, String p) { // dp[i][j] = p[0..i) matches s[0..j) // dp[i][j] = if p[i - 1] == s[j - 1] || p[j - 1] == '.' // dp[i - 1][j - 1] // if p[j - 1] == '*' // dp[i - 2][j] || dp[i][j - 1] && s[i - 1] == p[i - 2] || p[i - 2] == '.' // * matches 0 elements * matches s[i - 1] // dp[0][0] = true // dp[i][0] = true if dp[i - 2][0] and p[i - 1] == '*' // dp[m][n] int m = p.length(); int n = s.length(); boolean[][] dp = new boolean[m + 1][n + 1]; dp[0][0] = true; for (int i = 2; i <= m; i ++) { if (p.charAt(i - 1) == '*' && dp[i - 2][0]) { dp[i][0] = true; } } for (int i = 1; i <= m; i ++) { char r = p.charAt(i - 1); for (int j = 1; j <= n; j ++) { char c = s.charAt(j - 1); if (c == r || r == '.') { dp[i][j] = dp[i - 1][j - 1]; } else if (r == '*') { dp[i][j] = dp[i - 2][j] || dp[i][j - 1] && (c == p.charAt(i - 2) || p.charAt(i - 2) == '.'); } } } return dp[m][n]; } }