leetcode 212. Word Search II Add to List 查找单词---------- java

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.

 

 

1、使用及表暴力的算法，果然超时了。

public class Solution {
    
    public List<String> findWords(char[][] board, String[] words) {
        List<String> list = new ArrayList();
        for (String str : words){
            if (isExit(str, board)){
                if (!list.contains(str))
                    list.add(str);
            }
        }
        return list;
    }
    
    public boolean isExit(String word, char[][] board){
        char[] ch = word.toCharArray();
        char first = ch[0];
        for (int i = 0; i < board.length; i++){
            for (int j = 0; j < board[0].length; j++){
                if (board[i][j] == first ){
                    board[i][j] = 'A';
                    if (getExit(board, ch, i, j, 1)){
                        board[i][j] = first;
                        return true;
                    }
                    board[i][j] = first;
                }
            }
        }
        return false;
    }
    public boolean getExit(char[][] board, char[] ch, int i, int j, int pos){
        
        if (pos == ch.length){
            return true;
        }
        if (i - 1 >= 0 && board[i - 1][j] == ch[pos]){
            board[i - 1][j] = 'A';
            if (getExit(board, ch, i - 1, j, pos + 1)){
                board[i - 1][j] = ch[pos];
                return true;
            }
            board[i - 1][j] = ch[pos];
        }
        if (i + 1 < board.length && board[i + 1][j] == ch[pos]){
            board[i + 1][j] = 'A';
            if (getExit(board, ch, i + 1, j, pos + 1)){
                board[i + 1][j] = ch[pos];
                return true;
            }
            board[i + 1][j] = ch[pos];
        }
        if (j - 1 >= 0 && board[i][j - 1] == ch[pos]){
            board[i][j - 1] = 'A';
            if (getExit(board, ch, i, j - 1, pos + 1)){
                board[i][j - 1] = ch[pos];
                return true;
            }
            board[i][j - 1] = ch[pos];
        }
        if (j + 1 < board[0].length && board[i][j + 1] == ch[pos]){
            board[i][j + 1] = 'A';
            if (getExit(board, ch, i, j + 1, pos + 1)){
                board[i][j + 1] = ch[pos];
                return true;
            }
            board[i][j + 1] = ch[pos];
        }
        return false;
    }
    
}

 

 

2、利用前缀树，把单词建树，然后遍历一次所有节点开头的字符串即可。

public class Solution {
    
    public List<String> findWords(char[][] board, String[] words) {
    List<String> res = new ArrayList<>();
    TrieNode root = buildTrie(words);
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            dfs (board, i, j, root, res);
        }
    }
    return res;
}

public void dfs(char[][] board, int i, int j, TrieNode p, List<String> res) {
    char c = board[i][j];
    if (c == '#' || p.next[c - 'a'] == null) return;
    p = p.next[c - 'a'];
    if (p.word != null) {   // found one
        res.add(p.word);
        p.word = null;     // de-duplicate
    }

    board[i][j] = '#';
    if (i > 0) dfs(board, i - 1, j ,p, res); 
    if (j > 0) dfs(board, i, j - 1, p, res);
    if (i < board.length - 1) dfs(board, i + 1, j, p, res); 
    if (j < board[0].length - 1) dfs(board, i, j + 1, p, res); 
    board[i][j] = c;
}

public TrieNode buildTrie(String[] words) {
    TrieNode root = new TrieNode();
    for (String w : words) {
        TrieNode p = root;
        for (char c : w.toCharArray()) {
            int i = c - 'a';
            if (p.next[i] == null) p.next[i] = new TrieNode();
            p = p.next[i];
       }
       p.word = w;
    }
    return root;
}

class TrieNode {
    TrieNode[] next = new TrieNode[26];
    String word;
}
    
}