Implement Trie (Prefix Tree)
Implement a trie with insert
, search
, and startsWith
methods.
Note:
You may assume that all inputs are consist of lowercase letters a-z
solution:
class TrieNode {
// Initialize your data structure here.
boolean isEnd; //是否有单词以该字母结尾
TrieNode[] sons; //子节点[a-z]
char value; //保存该节点代表的字母值 public TrieNode() {
isEnd = false;
sons = new TrieNode[26];
}
} public class Trie {
private TrieNode root; public Trie() {
root = new TrieNode();
} // Inserts a word into the trie.
/**
* 对于传入的单词,从左往右遍历每个字母,依次从根节点向下建树
* 每个节点保存一个字母
* @param word
*/
public void insert(String word) {
if (word == null || word.length() == 0)
return;
TrieNode p = root; //工作指针,初始指向根节点
for (int i = 0; i < word.length(); i++) {
int pos = word.charAt(i) - 'a'; //hash-计算该字母对应的位置
if (p.sons[pos] == null) { //判断该位置是否为空
p.sons[pos] = new TrieNode();
p.sons[pos].value = word.charAt(i);
}
p = p.sons[pos]; //指向子节点
}
p.isEnd = true; //标记该节点,表示有单词以之结尾
} // Returns if the word is in the trie.
/**
* 类似建树过程,从左往右简历单词的每个字母,如果对应位置为空,则表示树中
* 不存在该单词
* @param word
* @return
*/
public boolean search(String word) {
if (word == null || word.length() == 0)
return false;
TrieNode p = root;
for (int i = 0; i < word.length(); i++) {
int pos = word.charAt(i) - 'a';
if (p.sons[pos] == null)
return false;
p = p.sons[pos];
}
return p.isEnd; //false表示所查单词是树中某个单词的前缀,但不完全匹配
} // Returns if there is any word in the trie
// that starts with the given prefix.
public boolean startsWith(String prefix) {
if (prefix == null || prefix.length() == 0)
return false;
TrieNode p = root;
for (int i = 0; i < prefix.length(); i++) {
int pos = prefix.charAt(i) - 'a';
if (p.sons[pos] == null)
return false;
p = p.sons[pos];
}
return true;
}
} // Your Trie object will be instantiated and called as such:
// Trie trie = new Trie();
// trie.insert("somestring");
// trie.search("key");
Add and Search Word
Design a data structure that supports the following two operations:
void addWord(word)
bool search(word)
search(word) can search a literal word or a regular expression string containing only letters a-z
or .
. A .
means it can represent any one letter.
For example:
addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true
Note:
You may assume that all words are consist of lowercase letters a-z
.
solution:
public class WordDictionary { private Tode root; public WordDictionary() {
root = new Tode();
} // Adds a word into the data structure.
/**
* 与上题一致
* @param word
*/
public void addWord(String word) {
if(word == null || word.length() == 0)
return ;
Tode p = root;
for(int i=0; i<word.length(); i++){
int pos = word.charAt(i) - 'a';
if(p.sons[pos] == null){
p.sons[pos] = new Tode();
p.sons[pos].value = word.charAt(i);
}
p = p.sons[pos];
}
p.isEnd = true;
} // Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
/**
* 因为.能匹配所有字母,导致不能直接hash判断字母是否存在,而必须对树进行深度优先遍历。
* @param word
* @return
*/
public boolean search(String word) {
if(word == null || word.length() == 0)
return false;
char[] arr = word.toCharArray();
int index = 0;
Tode p = root;
return goSearch(arr, index, p);
} private boolean goSearch(char[] arr, int index, Tode p) {
if(index == arr.length)
return p.isEnd;
if(arr[index] == '.'){
for(Tode n : p.sons){ //搜索当前节点的所有非空子节点
if(n != null && goSearch(arr, index+1, n)) //若有节点搜索到匹配单词,则直接返回
return true;
}
return false;
}else{
int pos = arr[index] - 'a';
if(p.sons[pos] != null)
return goSearch(arr, index+1, p.sons[pos]);
else
return false;
}
} class Tode{
boolean isEnd;
Tode[] sons;
char value;
public Tode(){
isEnd = false;
sons = new Tode[26];
}
}
} // Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");