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 precedeng 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
Problem link
You can find the detailed video tutorial here
The thought process is very similar to Leetcode Solution 44: Wildcard Matching, you can find the blog here and the video tutorial here. Step 0 really is to understand what the "*" really means.
'*' Matches zero or more of the preceding element.
A few examples,
The first attempt (if you haven't seen this question before) is trying to solve it using recursion. The idea is simple. While comparing the s and p char by char, as soon as you see a "*", you will start recurse
The time complexity in a ballpark looks like exponential to the P length, but a more detailed analysis from leetcode's original solution can be found here. If you recall what problems can be solved using DP, DP (Dynamic Programming) is normally a great way when trying to answer below two types of questions, and string related questions are a majority use case of DP
I have posted several videos on how to solve DP problems using a template, essentially
This problem can be solved using DP by having formulas like below when a "*" is see
Once you have the formula, coding up DP problem should be a piece of cake. You can find the video tutorial on this problem here
public boolean isMatch(String s, String p) { | |
---|---|
if (s == null || p == null) { | |
return false; | |
} | |
int row = s.length(); | |
int col = p.length(); | |
boolean[][] lookup = new boolean[row + 1][col + 1]; | |
lookup[0][0] = true; | |
// init the init values | |
for (int j = 1; j <= col; j++) { | |
if (p.charAt(j - 1) == '*') { | |
if (lookup[0][j - 1]) { | |
lookup[0][j] = true; | |
} else if (j >= 2 && lookup[0][j - 2]) { | |
lookup[0][j] = true; | |
} | |
} | |
} | |
for (int i = 1; i <= row; i++) { | |
for (int j = 1; j <= col; j++) { | |
if (p.charAt(j - 1) == '*') { | |
// in the graph above referred as first if | |
if (lookup[i][j - 1]) { | |
lookup[i][j] = true; | |
} | |
// in the graph above referred as second if | |
if (!lookup[i][j] && lookup[i - 1][j]) { | |
lookup[i][j] = (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.'); | |
} | |
// in the graph above referred as third if | |
if (!lookup[i][j] && j >= 2 && lookup[i][j - 2]) { | |
lookup[i][j] = true; | |
} | |
} else { | |
if (lookup[i - 1][j - 1] && ((p.charAt(j - 1) == s.charAt(i - 1)) || p.charAt(j - 1) == '.')) { | |
lookup[i][j] = true; | |
} | |
} | |
} | |
} | |
// Utilities.printBooleanMatrix(lookup); | |
return lookup[row][col]; | |
} |
view rawRegularExpresionMatching.java hosted with ❤ by GitHub
Time Complexity: assuming S length is M, P length is N, this is O(M*N) because all we need to do is build up that lookup matrix Space Complexity: assuming S length is M, P length is N, this is O(M*N), again, that lookup matrix