blob: fd783c5929ec8a198201049fb4b0bc692e963d89 [file] [log] [blame]
// Copyright 2014 The Bazel Authors. All rights reserved.
//
// 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.google.devtools.build.lib.vfs;
import com.google.common.base.Preconditions;
import com.google.devtools.build.lib.concurrent.ThreadSafety.ThreadCompatible;
import java.util.HashMap;
import java.util.Map;
/**
* A trie that operates on path segments.
*
* @param <T> the type of the values.
*/
@ThreadCompatible
public class PathTrie<T> {
@SuppressWarnings("unchecked")
private static class Node<T> {
private Node() {
children = new HashMap<>();
}
private T value;
private Map<String, Node<T>> children;
}
private final Node<T> root;
public PathTrie() {
root = new Node<T>();
}
/**
* Puts a value in the trie.
*
* @param key must be an absolute path.
*/
public void put(PathFragment key, T value) {
Preconditions.checkArgument(key.isAbsolute(), "PathTrie only accepts absolute paths as keys.");
Node<T> current = root;
for (String segment : key.getSegments()) {
current.children.putIfAbsent(segment, new Node<T>());
current = current.children.get(segment);
}
current.value = value;
}
/**
* Gets a value from the trie. If there is an entry with the same key, that will be returned,
* otherwise, the value corresponding to the key that matches the longest prefix of the input.
*/
public T get(PathFragment key) {
Node<T> current = root;
T lastValue = current.value;
for (String segment : key.getSegments()) {
if (current.children.containsKey(segment)) {
current = current.children.get(segment);
// Track the values of increasing matching prefixes.
if (current.value != null) {
lastValue = current.value;
}
} else {
// We've reached the longest prefix, no further to go.
break;
}
}
return lastValue;
}
}