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"].

Note:
You may assume that all inputs are consist of lowercase letters a-z.

click to show hint.

You would need to optimize your backtracking to pass the larger test. Could you stop backtracking earlier?

If the current candidate does not exist in all words' prefix, you could stop backtracking immediately. What kind of data structure could answer such query efficiently? Does a hash table work? Why or why not? How about a Trie? If you would like to learn how to implement a basic trie, please work on this problem: Implement Trie (Prefix Tree) first.

这是一道DFS和TrieTree结合的经典题。首先要实现一个TrieTree，把words加入TrieTree里，然后用DFS遍历整个数组，沿着TrieTree查找单词，一旦发现，加入到res中。代码如下：

public class Solution {    class TrieNode {        String s;        boolean isString;        HashMap<Character, TrieNode> children = new HashMap<Character, TrieNode>();        public TrieNode () {            this.isString = false;        }    };        class TrieTree {        TrieNode root;        public TrieTree() {            root = new TrieNode();        }        public void insert(String s) {            TrieNode currNode = root;            char[] chs = s.toCharArray();            for (int i = 0; i < chs.length; i ++) {                if (!currNode.children.containsKey(chs[i])) {                    currNode.children.put(chs[i], new TrieNode());                }                 currNode = currNode.children.get(chs[i]);            }            currNode.s = s;            currNode.isString = true;        }        public boolean search(String s) {            TrieNode currNode = root;            char[] chs = s.toCharArray();            for (int i = 0; i < chs.length; i ++) {                if (!currNode.children.containsKey(chs[i])) {                    return false;                }                 currNode = currNode.children.get(chs[i]);            }            return currNode.isString;        }    };        boolean[][] visited;    int[][] dirs = new int[][]{{1,0}, {-1,0}, {0,1}, {0,-1}};        public List<String> findWords(char[][] board, String[] words) {        List<String> res = new ArrayList<String>();        if (board == null || board.length == 0) {            return res;        }        int m = board.length, n = board[0].length;        visited = new boolean[m][n];        TrieTree tree = new TrieTree();        for (String word: words) {            tree.insert(word);        }        for (int i = 0; i < m; i ++) {            for (int j = 0; j < n; j ++) {                DFS(board, i, j, tree.root, res);            }        }        return res;    }        private void DFS(char[][] board, int i, int j, TrieNode node, List<String> list) {        if (node.isString == true) {            if (!list.contains(node.s)) {                list.add(node.s);            }        }        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || visited[i][j] || node == null) {            return;        }        if (node.children.containsKey(board[i][j])) {            visited[i][j] = true;            for (int[] dir: dirs) {                DFS(board, i + dir[0], j + dir[1], node.children.get(board[i][j]), list);            }            visited[i][j] = false;        }    }}