LeetCode - Regular Expression Matching
Problem description
Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.
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'.' 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:
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Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Example 2:
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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:
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Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
Example 4:
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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:
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Input:
s = "mississippi"
p = "mis*is*p*."
Output: false
Analysis
Basic idea is to use recursion and substring to make this problem into simple one.
Considering the '*', there are several situations.
aa matches a*, we can convert it into a matches a* if the first character is matched.
Another situation is ba matches c*ba, we should skip the c*
Here’s the code:
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public class RegularExpressionMatching {
public boolean isMatch(String s, String p) {
if (p.isEmpty()) {
return s.isEmpty();
}
boolean isFirstMatched = !s.isEmpty() && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.');
if (p.length() >= 2 && p.charAt(1) == '*') {
return isMatch(s, p.substring(2))
|| (isFirstMatched && isMatch(s.substring(1), p));
} else {
return isFirstMatched && isMatch(s.substring(1), p.substring(1));
}
}
}
This solution is slow, and the best solution from LeetCode is:
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class Solution {
public boolean isMatch(String s, String p) {
boolean[][] T = new boolean[s.length() + 1][p.length() + 1];
T[0][0] = true;
for (int j = 1; j < T[0].length; j++) {
if (p.charAt(j - 1) == '*') {
T[0][j] = T[0][j - 2];
}
}
for (int i = 1; i < T.length; i++) {
for (int j = 1; j < T[0].length; j++) {
if (p.charAt(j - 1) == '.' || p.charAt(j - 1) == s.charAt(i - 1)) {
T[i][j] = T[i - 1][j - 1];
} else if (p.charAt(j - 1) == '*') {
T[i][j] = T[i][j - 2];
if (p.charAt(j - 2) == '.' || p.charAt(j - 2) == s.charAt(i - 1)) {
T[i][j] = T[i][j] | T[i - 1][j];
}
} else {
T[i][j] = false;
}
}
}
return T[s.length()][p.length()];
}
}
Basically it uses a 2D array to save all the information, but the logic is the same.