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"].

 

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

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.

public static final int[][] DIR = {{1,0}, {-1,0}, {0,1}, {0,-1}};
public static class TrieNode{
    TrieNode[] children = new TrieNode[26];
    boolean isLeaf;
}
public void insert(TrieNode node, String word, int pos) {
    if(pos >= word.length()) {
        node.isLeaf = true;
        return;
    }
    int index = word.charAt(pos)-'a';
    if(node.children[index] == null) {
        node.children[index] = new TrieNode();
    }
    insert(node.children[index], word, pos+1);
}
public void check(char[][] board, TrieNode node, int row, int col, StringBuilder sb, Set<String> list) {
    int m = board.length, n = board[0].length;
    if(row>=m || row<0 || col>=n || col<0 || board[row][col] == '.') return;
    char c = board[row][col];
    board[row][col] = '.';
    TrieNode child = node.children[c - 'a'];
    if(child != null) {
        sb.append(c);
        if(child.isLeaf) 
            list.add(sb.toString());
        for(int i=0; i<DIR.length; i++)
            check(board, child, row+DIR[i][0], col+DIR[i][1], sb, list);
        sb.deleteCharAt(sb.length()-1);
    }
    board[row][col] = c;
}
public List<String> findWords(char[][] board, String[] words) {
    Set<String> set = new HashSet<>();
    int m = board.length, n = board[0].length;
    TrieNode root = new TrieNode();
    for(String word:words) { // build trie from words
        insert(root, word, 0);
    }
    StringBuilder sb = new StringBuilder();
    for(int i=0; i<m; i++) {
        for(int j=0; j<n && set.size()<words.length; j++) {
            check(board, root, i, j, sb, set);
        }
    }
    return new ArrayList<>(set);
}

  

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转载自yuanhsh.iteye.com/blog/2217976