Implement a trie with insert, search, and startsWith methods. Note:
You may assume that all inputs are consist of lowercase letters a-z.
参考百度百科：Trie树
a trie, also called digital tree and sometimes radix tree or prefix tree (as they can be searched by prefixes)
The time complexity to insert and to search is O(m), where m is the length of the string.
标准Trie树的应用和优缺点
(1) 全字匹配：确定待查字串是否与集合的一个单词完全匹配。如上代码fullMatch()。
(2) 前缀匹配：查找集合中与以s为前缀的所有串。
注意：Trie树的结构并不适合用来查找子串。这一点和前面提到的PAT Tree以及后面专门要提到的Suffix Tree的作用有很大不同。
优点： 查找效率比与集合中的每一个字符串做匹配的效率要高很多。在o(m)时间内搜索一个长度为m的字符串s是否在字典里。Predictable O(k) lookup time where k is the size of the key
缺点：标准Trie的空间利用率不高，可能存在大量结点中只有一个子结点，这样的结点绝对是一种浪费。正是这个原因，才迅速推动了下面所讲的压缩trie的开发。
什么时候用Trie?
It all depends on what problem you're trying to solve. If all you need to do is insertions and lookups, go with a hash table. If you need to solve more complex problems such as prefix-related queries, then a trie might be the better solution.
像word search II就是跟前缀有关，如果dfs发现当前形成的前缀都不在字典中，就没必要再搜索下去了，所以用trie不用hashSet
Easy version of implement Trie. TrieNode only contains TrieNode[] children, and boolean isWord two fields
class Trie {
class TrieNode {
TrieNode[] children;
boolean isWord;
public TrieNode() {
this.children = new TrieNode[26];
this.isWord = false;
}
}
TrieNode root;
/** Initialize your data structure here. */
public Trie() {
this.root = new TrieNode();
}
/** Inserts a word into the trie. */
public void insert(String word) {
if (word == null || word.length() == 0) return;
TrieNode cur = this.root;
for (int i = 0; i < word.length(); i ++) {
if (cur.children[word.charAt(i) - 'a'] == null) {
cur.children[word.charAt(i) - 'a'] = new TrieNode();
}
cur = cur.children[word.charAt(i) - 'a'];
}
cur.isWord = true;
}
/** Returns if the word is in the trie. */
public boolean search(String word) {
TrieNode cur = this.root;
for (int i = 0; i < word.length(); i ++) {
if (cur.children[word.charAt(i) - 'a'] == null) return false;
cur = cur.children[word.charAt(i) - 'a'];
}
return cur.isWord;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
public boolean startsWith(String prefix) {
TrieNode cur = this.root;
for (int i = 0; i < prefix.length(); i ++) {
if (cur.children[prefix.charAt(i) - 'a'] == null) return false;
cur = cur.children[prefix.charAt(i) - 'a'];
}
return true;
}
}
Older version, TrieNode also has num and val fields, which might not be that useful.
class TrieNode {
// Initialize your data structure here.
int num; //How many words go through this TrieNode
TrieNode[] son; //collection of sons
boolean isEnd;
char val;
public TrieNode() {
this.num = 0;
this.son = new TrieNode[26];
this.isEnd = false;
}
}
public class Trie {
private TrieNode root;
public Trie() {
root = new TrieNode();
}
// Inserts a word into the trie.
public void insert(String word) {
if (word==null || word.length()==0) return;
char[] arr = word.toCharArray();
TrieNode node = this.root;
for (int i=0; i<arr.length; i++) {
int pos = (int)(arr[i] - 'a');
if (node.son[pos] == null) {
node.son[pos] = new TrieNode();
node.son[pos].num++;
node.son[pos].val = arr[i];
}
else {
node.son[pos].num++;
}
node = node.son[pos];
}
node.isEnd = true;
}
// Returns if the word is in the trie.
public boolean search(String word) {
char[] arr = word.toCharArray();
TrieNode node = this.root;
for (int i=0; i<arr.length; i++) {
int pos = (int)(arr[i] - 'a');
if (node.son[pos] == null) return false;
node = node.son[pos];
}
return node.isEnd;
}
// Returns if there is any word in the trie
// that starts with the given prefix.
public boolean startsWith(String prefix) {
char[] arr = prefix.toCharArray();
TrieNode node = this.root;
for (int i=0; i<arr.length; i++) {
int pos = (int)(arr[i] - 'a');
if (node.son[pos] == null) return false;
node = node.son[pos];
}
return true;
}
}
// Your Trie object will be instantiated and called as such:
// Trie trie = new Trie();
// trie.insert("somestring");
// trie.search("key");