Java for LeetCode 212 Word Search II

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

For example,
Given words = ["oath","pea","eat","rain"] and board =

[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]

Return ["eat","oath"].

解题思路：

本题承接自Java for LeetCode 079 Word Search 但是用dfs会TLE，真正的做法是将word放进Trie里面，然后遍历trie里的每个点，看是否能匹配到trie里的元素，JAVA实现如下：

public class Solution {
	public List<String> findWords(char[][] board, String[] words) {
		HashSet<String> list=new HashSet();
		Trie trie = new Trie();
        for (String word : words)
            trie.insert(word);
        boolean[][] visited=new boolean[board.length][board[0].length];
        for (int i = 0; i < board.length; i++)
            for (int j = 0; j < board[0].length; j++)
                dfs(list,board, visited, "", i, j, trie);
        return new ArrayList(list);
    }  

    public void dfs(Set<String> list,char[][] board, boolean[][] visited, String str, int x, int y, Trie trie) {
        if (x < 0 || x >= board.length || y < 0 || y >= board[0].length)
        	return;
        if (visited[x][y])
        	return;
        str += board[x][y];
        if (!trie.startsWith(str))
        	return;
        if (trie.search(str))
            list.add(str);
        visited[x][y] = true;
        dfs(list,board, visited, str, x - 1, y, trie);
        dfs(list,board, visited, str, x + 1, y, trie);
        dfs(list,board, visited, str, x, y - 1, trie);
        dfs(list,board, visited, str, x, y + 1, trie);
        visited[x][y] = false;
    }
}
class TrieNode {
	// Initialize your data structure here.
	int num;// 有多少单词通过这个节点,即节点字符出现的次数
	TrieNode[] son;// 所有的儿子节点
	boolean isEnd;// 是不是最后一个节点
	char val;// 节点的值

	TrieNode() {
		this.num = 1;
		this.son = new TrieNode[26];
		this.isEnd = false;
	}
}

class Trie {
	protected TrieNode root;

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

	public void insert(String word) {
		if (word == null || word.length() == 0)
			return;
		TrieNode node = this.root;
		char[] letters = word.toCharArray();
		for (int i = 0; i < word.length(); i++) {
			int pos = letters[i] - 'a';
			if (node.son[pos] == null) {
				node.son[pos] = new TrieNode();
				node.son[pos].val = letters[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) {
		if (word == null || word.length() == 0) {
			return false;
		}
		TrieNode node = root;
		char[] letters = word.toCharArray();
		for (int i = 0; i < word.length(); i++) {
			int pos = letters[i] - 'a';
			if (node.son[pos] != null) {
				node = node.son[pos];
			} else {
				return false;
			}
		}
		return node.isEnd;
	}

	// 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 node = root;
		char[] letters = prefix.toCharArray();
		for (int i = 0; i < prefix.length(); i++) {
			int pos = letters[i] - 'a';
			if (node.son[pos] != null) {
				node = node.son[pos];
			} else {
				return false;
			}
		}
		return true;
	}
}