Regular Expression Matching
描述
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).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true
分析
这是一道很有挑战的题。
递归版
// Regular Expression Matching
// Time complexity: O(n)
// Space complexity: O(1)
class Solution {
public boolean isMatch(final String s, final String p) {
return isMatch(s, 0, p, 0);
}
private static boolean matchFirst(String s, int i, String p, int j) {
if (j == p.length()) return i == s.length();
if (i == s.length()) return j == p.length();
return p.charAt(j) == '.' || s.charAt(i) == p.charAt(j);
}
private static boolean isMatch(String s, int i, String p, int j) {
if (j == p.length()) return i == s.length();
// next char is not '*', then must match current character
final char b = p.charAt(j);
if (j == p.length() - 1 || p.charAt(j + 1) != '*') {
if (matchFirst(s, i, p, j)) return isMatch(s, i + 1, p, j + 1);
else return false;
} else { // next char is '*'
if (isMatch(s, i, p, j+2)) return true; // try the length of 0
while (matchFirst(s, i, p, j)) // try all possible lengths
if (isMatch(s, ++i, p, j+2)) return true;
return false;
}
}
}