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SimpleRegularExpression.java
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SimpleRegularExpression.java
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/*
* Copyright (C) 2014 Pedro Vicente Gómez Sánchez.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package com.github.pedrovgs.problem33;
/**
* Can you write a method to analyze some strings with regular expressions and return true if the
* expression matches with the word? Expressions supported are:
* '.' Matches any single character.
* '*' Matches zero or more of the preceding element.
*
* @author Pedro Vicente Gómez Sánchez.
*/
public class SimpleRegularExpression {
/**
* Tail recursive solution to this problem. The complexity order of this algorithm is O(N^2) in
* time terms and O(1) in space terms because we are not using any additional data structure to
* solve this problem.
*/
public boolean evaluate(String source, String pattern) {
if (source == null || pattern == null) {
throw new IllegalArgumentException("You can't pass a null strings as input");
}
if (pattern.length() == 0) return source.length() == 0;
// Length 1 is special case
if (pattern.length() == 1 || pattern.charAt(1) != '*') {
if (source.length() < 1 || (pattern.charAt(0) != '.' && source.charAt(0) != pattern.charAt(
0))) {
return false;
}
return evaluate(source.substring(1), pattern.substring(1));
} else {
int len = source.length();
int i = -1;
while (i < len && (i < 0 || pattern.charAt(0) == '.' || pattern.charAt(0) == source.charAt(
i))) {
if (evaluate(source.substring(i + 1), pattern.substring(2))) return true;
i++;
}
return false;
}
}
}