208. Implement Trie (Prefix Tree)
原题链接:https://leetcode.com/problems/implement-trie-prefix-tree/description/
都是前缀树相关的题目。LeetCode 上面刷题时可以查看 Similar Questions 问题,这道题目也是做 211. Add and Search Word - Data structure design 这道题目看到的。
这道题目就很简单的,我之前已经专门看过了前缀树的实现了:http://www.cnblogs.com/optor/p/8513547.html
所以,这道题就是直接把之前的代码复制过来就好了:
class Trie {
class TrieNode {
// R links to node children
private TrieNode[] links;
private final int R = 26;
private boolean isEnd;
// number of children non null links
private int size;
public TrieNode() {
links = new TrieNode[R];
}
public boolean containsKey(char ch) {
return links[ch - 'a'] != null;
}
public TrieNode get(char ch) {
return links[ch - 'a'];
}
public void put(char ch, TrieNode node) {
links[ch - 'a'] = node;
size++;
}
public int getLinks() {
return size;
}
public void setEnd() {
isEnd = true;
}
public boolean isEnd() {
return isEnd;
}
}
private TrieNode root;
/** Initialize your data structure here. */
public Trie() {
root = new TrieNode();
}
/** Inserts a word into the trie. */
public void insert(String word) {
TrieNode node = root;
for (int i = 0; i < word.length(); i++) {
char currentChar = word.charAt(i);
if (!node.containsKey(currentChar)) {
node.put(currentChar, new TrieNode());
}
node = node.get(currentChar);
}
node.setEnd();
}
private TrieNode searchPrefix(String word) {
TrieNode node = root;
for (int i = 0; i < word.length(); i++) {
char curLetter = word.charAt(i);
if (node.containsKey(curLetter)) {
node = node.get(curLetter);
} else {
return null;
}
}
return node;
}
/** Returns if the word is in the trie. */
public boolean search(String word) {
TrieNode node = searchPrefix(word);
return node != null && node.isEnd();
}
/** Returns if there is any word in the trie that starts with the given prefix. */
public boolean startsWith(String prefix) {
TrieNode node = searchPrefix(prefix);
return node != null;
}
}
/**
* Your Trie object will be instantiated and called as such:
* Trie obj = new Trie();
* obj.insert(word);
* boolean param_2 = obj.search(word);
* boolean param_3 = obj.startsWith(prefix);
*/