LeetCode - Word Search

問題の説明

Given an m x n grid of characters board and a string word, return true if word exists in the grid.

The word can 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.

I/O例

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" Output: true

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" Output: true

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" Output: false

せいげんじょうけん

m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board and word consists of only lowercase and uppercase English letters.

Solution

import java.util.Arrays;

class Solution {
    public boolean exist(char[][] board, String word) {

        boolean[][] visits = new boolean[board.length][board[0].length];

        for (boolean[] visit : visits) {
            Arrays.fill(visit, false);
        }

        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[i].length; j++) {
                // System.out.println("i , j :"+ i+ " : "+j);
                if (backtracking(board, visits, i, j, word, 0)) {
                    return true;
                }
            }
        }
        return false;
    }

    public boolean backtracking(char[][] board, boolean[][] visits, int i, int j, String word, int len) {

        if (board[i][j] == word.charAt(len)) {

            if (word.length() == len + 1) {
                return true;
            }
            visits[i][j] = true;
            boolean up = i <= 0 || visits[i - 1][j] ? false : backtracking(board, visits, i - 1, j, word, len + 1);
            boolean down = i >= board.length - 1 || visits[i + 1][j] ? false
                    : backtracking(board, visits, i + 1, j, word, len + 1);
            boolean left = j <= 0 || visits[i][j - 1] ? false : backtracking(board, visits, i, j - 1, word, len + 1);
            boolean right = j >= board[0].length - 1 || visits[i][j + 1] ? false
                    : backtracking(board, visits, i, j + 1, word, len + 1);

            visits[i][j] = false;
            return up || down || left || right;
        }
        return false;
    }
}