Given string s and pattern p with . and * support, implement regular expression matching. . matches any single character, * matches zero or more of the preceding element.
Input: s="aa", p="a*" → Output: trueInput: s="ab", p=".*" → Output: true
dp[i][j] = whether s[0..i-1] matches p[0..j-1]. Handle * by either skipping the preceding pattern element or consuming one character from s.
- dp[0][0] = true; dp[0][j] = true if pattern[j-1]=="*" and dp[0][j-2].
- If p[j-1]==s[i-1] or p[j-1]==".": dp[i][j] = dp[i-1][j-1].
- If p[j-1]=="*": zero occurrences dp[i][j-2], or one+ occurrences dp[i-1][j] if p[j-2] matches s[i-1].
class Solution {
public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int j = 2; j <= n; j += 2)
if (p.charAt(j-1) == '*') dp[0][j] = dp[0][j-2];
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (p.charAt(j-1) == '*') {
dp[i][j] = dp[i][j-2]; // zero
if (p.charAt(j-2) == '.' || p.charAt(j-2) == s.charAt(i-1))
dp[i][j] = dp[i][j] || dp[i-1][j]; // one+
} else if (p.charAt(j-1) == '.' || p.charAt(j-1) == s.charAt(i-1)) {
dp[i][j] = dp[i-1][j-1];
}
}
}
return dp[m][n];
}
}
- Time Complexity: O(M*N)
- Space Complexity: O(M*N)