Leetcode 精选之搜索（单词搜索）

题目描述

给定一个二维网格和一个单词，找出该单词是否存在于网格中。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例:

board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

给定 word = "ABCCED", 返回 true
给定 word = "SEE", 返回 true
给定 word = "ABCB", 返回 false

提示：

• board 和 word 中只包含大写和小写英文字母。
• 1 <= board.length <= 200
• 1 <= board[i].length <= 200
• 1 <= word.length <= 10^3

题目链接：力扣

解题思路

class Solution {
    private final static int[][] direction = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    private int m;
    private int n;

    public boolean exist(char[][] board, String word) {
        if (word == null || word.length() == 0) {
            return true;
        }
        if (board == null || board.length == 0 || board[0].length == 0) {
            return false;
        }

        m = board.length;
        n = board[0].length;
        boolean[][] hasVisited = new boolean[m][n];

        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (backtracking(0, r, c, hasVisited, board, word)) {
                    return true;
                }
            }
        }

        return false;
    }

    private boolean backtracking(int curLen, int r, int c, boolean[][] visited, final char[][] board, final String word) {
        if (curLen == word.length()) {
            return true;
        }
        if (r < 0 || r >= m || c < 0 || c >= n
                || board[r][c] != word.charAt(curLen) || visited[r][c]) {

            return false;
        }

        visited[r][c] = true;

        for (int[] d : direction) {
            if (backtracking(curLen + 1, r + d[0], c + d[1], visited, board, word)) {
                return true;
            }
        }

        visited[r][c] = false;

        return false;
    }
}

测试结果

image.png