LeetCode-in-Java

212. Word Search II

Hard

Given an m x n board of characters and a list of strings words, return all words on the board.

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

Example 1:

Input: board = [[“o”,”a”,”a”,”n”],[“e”,”t”,”a”,”e”],[“i”,”h”,”k”,”r”],[“i”,”f”,”l”,”v”]], words = [“oath”,”pea”,”eat”,”rain”]

Output: [“eat”,”oath”]

Example 2:

Input: board = [[“a”,”b”],[“c”,”d”]], words = [“abcb”]

Output: []

Constraints:

• m == board.length
• n == board[i].length
• 1 <= m, n <= 12
• board[i][j] is a lowercase English letter.
• 1 <= words.length <= 3 * 104
• 1 <= words[i].length <= 10
• words[i] consists of lowercase English letters.
• All the strings of words are unique.

Solution

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Solution {
    private Tree root;

    public List<String> findWords(char[][] board, String[] words) {
        if (board.length < 1 || board[0].length < 1) {
            return Collections.emptyList();
        }
        root = new Tree();
        for (String word : words) {
            Tree.addWord(root, word);
        }
        List<String> collected = new ArrayList<>();
        for (int i = 0; i != board.length; ++i) {
            for (int j = 0; j != board[0].length; ++j) {
                dfs(board, i, j, root, collected);
            }
        }
        return collected;
    }

    private void dfs(char[][] board, int i, int j, Tree cur, List<String> collected) {
        char c = board[i][j];
        if (c == '-') {
            return;
        }
        cur = cur.getChild(c);
        if (cur == null) {
            return;
        }
        if (cur.end != null) {
            String s = cur.end;
            collected.add(s);
            cur.end = null;
            if (cur.len() == 0) {
                Tree.deleteWord(root, s);
            }
        }
        board[i][j] = '-';
        if (i > 0) {
            dfs(board, i - 1, j, cur, collected);
        }
        if (i + 1 < board.length) {
            dfs(board, i + 1, j, cur, collected);
        }
        if (j > 0) {
            dfs(board, i, j - 1, cur, collected);
        }
        if (j + 1 < board[0].length) {
            dfs(board, i, j + 1, cur, collected);
        }
        board[i][j] = c;
    }
}

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