# Word Search II

Given a matrix of lower alphabets and a dictionary. Find all words in the dictionary that can be found in the matrix. A word can start from any position in the matrix and go left/right/up/down to the adjacent position.

**Example**

Given matrix:

```
doaf
agai
dcan
```

and dictionary:

```
{"dog", "dad", "dgdg", "can", "again"}
```

return {"dog", "dad", "can", "again"}

分析：

初始就判断valid,因为叶节点也没必要继续扩展下去。同时剪枝是在for loop外，为了一个元素的时候也能判断。

```
public class Solution {
    /**
     * @param board: A list of lists of character
     * @param words: A list of string
     * @return: A list of string
     */
     class TrieNode {
        String s;
         boolean isString;
         HashMap<Character, TrieNode> subtree;
         public TrieNode() {
            // TODO Auto-generated constructor stub
             isString = false;
             subtree = new HashMap<Character, TrieNode>();
             s = "";
         }
    };


    class TrieTree{
        TrieNode root ;
        public TrieTree(TrieNode TrieNode) {
            root = TrieNode;
        }
        public void insert(String s) {
            TrieNode now = root;
            for (int i = 0; i < s.length(); i++) {
                if (!now.subtree.containsKey(s.charAt(i))) {
                    now.subtree.put(s.charAt(i), new TrieNode());
                }
                now  =  now.subtree.get(s.charAt(i));
            }
            now.s = s;
            now.isString  = true;
        }
        public boolean find(String s){
            TrieNode now = root;
            for (int i = 0; i < s.length(); i++) {
                if (!now.subtree.containsKey(s.charAt(i))) {
                    return false;
                }
                now  =  now.subtree.get(s.charAt(i));
            }
            return now.isString ;
        }
    };

     ArrayList<String> ret;
     char[][] board;
     int n,m;
     int[] dx = {1,-1,0,0};
     int[] dy = {0,0,1,-1};

     public void search(int x, int y, TrieNode root){

         if(root.isString == true){
             if(!ret.contains(root.s)){
                 ret.add(root.s);
             }
         }
//只能在此判断是为了只有一个元素的时候，在for里判断直接continue不会进行下一轮dfs。这里初始就判断是否valid.
         if(x < 0 || x >= n || y < 0 || y >= m || board[x][y]==0 || root ==null)
            return;

        if(root.subtree.containsKey(board[x][y])){
            for(int i = 0; i < 4; i ++){
                char temp = board[x][y];
                board[x][y] = 0;
                search(x+dx[i], y+dy[i], root.subtree.get(temp));
                board[x][y] = temp;
            }
        }
     }

    public ArrayList<String> wordSearchII(char[][] board, ArrayList<String> words) {
        // write your code here

        this.ret = new ArrayList<String>();
        this.board = board;
        this.n = board.length;
        this.m = board[0].length;

        TrieTree tree = new TrieTree(new TrieNode());
        for(String word : words){
            tree.insert(word);
        }

        for(int i = 0; i< board.length; i++){
            for(int j=0; j< board[0].length; j++){
                search(i, j, tree.root);

            }
        }
        return ret;

    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://nataliekung.gitbook.io/ladder_code/qiang-hua-2-data-structure-i/word-search-ii.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.