leetcode-trie树实现

题目大意

　　https://leetcode.com/problems/implement-trie-prefix-tree/

　　trie树的实现

题目分析

　　trie树，实现插入，搜索，复杂度都是o(len),len为单词的长度

代码

• Java版本

class TrieNode {
       
    public TrieNode[] childNodes;
    public int freq;
    public char nodeChar;
    // Initialize your data structure here.
    public TrieNode() {
        childNodes = new TrieNode[26];
        freq = 0;
    }
}

public class Trie {
    private TrieNode root;

    public Trie() {
        root = new TrieNode();
    }

    // Inserts a word into the trie.
    public void insert(String word) {
        if(word.length() == 0) {
            return;
        }
        TrieNode cur = root;
        for(int i = 0; i < word.length(); i++) {
            int k = word.charAt(i) - 'a';
            if(cur.childNodes[k] == null) {
            	cur.childNodes[k] = new TrieNode();
            	cur.childNodes[k].nodeChar = word.charAt(i);
            }
            cur = cur.childNodes[k];
            if(i == word.length() - 1) {
            	cur.freq ++;
            }
        }
    }

    // Returns if the word is in the trie.
    //root->'a'->'b'->'c'
    public boolean search(String word) {
        if(word.length() == 0) {
        	return false;
        }
        TrieNode cur = root;
        for(int i = 0; i < word.length(); i++) {
            int k = word.charAt(i) - 'a';
            if(cur.childNodes[k] == null) {
                return false;
            }
            cur = cur.childNodes[k];
        }
        if(cur.freq > 0) {
        	return true;
        } else {
        	return false;
        }
    }

    // Returns if there is any word in the trie
    // that starts with the given prefix.
    public boolean startsWith(String prefix) {
    	if(prefix.length() == 0) {
        	return false;
        }
        TrieNode cur = root;
        for(int i = 0; i < prefix.length(); i++) {
            int k = prefix.charAt(i) - 'a';
            if(cur.childNodes[k] == null) {
                return false;
            }
            cur = cur.childNodes[k];
        }
        return true;
    }
}

－ C++版本

class TrieNode {
public:
    // Initialize your data structure here.
    char nodeChar;
    int freq;
    TrieNode* childNodes[26];
    TrieNode() {
        for(int i = 0; i < 26; i++) {
            childNodes[i] = NULL;
        }
        freq = 0;
    }
};

class Trie {
public:
    Trie() {
        root = new TrieNode();
    }
    
    // Inserts a word into the trie.
    void insert(string word) {
        int s = word.length();
        if(s == 0) {
            return;
        }
        TrieNode* cur = root;
        for(int i = 0; i < s; i++) {
            int k = word[i] - 'a';
            if((cur->childNodes)[k] == NULL) {
                (cur->childNodes)[k] = new TrieNode();
                (cur->childNodes)[k]->nodeChar = word[i];
            }
            cur = (cur->childNodes)[k];
            if(i == s - 1) {
                cur->freq ++;
            }
        }
    }
    
    // Returns if the word is in the trie.
    bool search(string word) {
        int s = word.length();
        if(s == 0) {
            return false;
        }
        TrieNode* cur = root;
        for(int i = 0; i < s; i++) {
            int k = word[i] - 'a';
            if((cur->childNodes)[k] == NULL) {
                return false;
            }
            cur = (cur->childNodes)[k];
        }
        if(cur->freq > 0) {
            return true;
        } else {
            return false;
        }
    }
    
    // Returns if there is any word in the trie
    // that starts with the given prefix.
    bool startsWith(string prefix) {
        int s = prefix.length();
        if(s == 0) {
            return false;
        }
        TrieNode* cur = root;
        for(int i = 0; i < s; i++) {
            int k = prefix[i] - 'a';
            if((cur->childNodes)[k] == NULL) {
                return false;
            }
            cur = (cur->childNodes)[k];
        }
        return true;
    }
    
private:
    TrieNode* root;
};