79. Word Search

class Solution {
    private static final int[][] DIRECTIONS = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    public boolean exist(char[][] board, String word) {
        // 注意每个点都能作为搜索起点
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (dfs(board, i, j, word, 0)) {
                    return true;
                }
            }
        }

        return false;
    }

    private boolean dfs(char[][] board, int i, int j, String word, int index) {
        int m = board.length, n = board[0].length;

        if (index == word.length()) {
            return true;
        }

        if (board[i][j] != word.charAt(index)) {
            return false;
        }

        if (index == word.length() - 1) {
            // 处理二维网格只有一个字符的场景，board: [["a"]], word: "a" 
            return true;
        }

        char c = board[i][j];
        board[i][j] = ' ';

        for (int[] direction : DIRECTIONS) {
            int x = i + direction[0];
            int y = j + direction[1];
            if (x >= 0 && x < m && y >= 0 && y < n && dfs(board, x, y, word, index + 1)) {
                return true;
            }
        }

        board[i][j] = c;
        return false;
    }
}

以上解法在能够搜索到单词的情况下修改了原二维网格，如果原二维网格不能修改则需要做另行处理。

Reference

79. Word Search
剑指 Offer 12. 矩阵中的路径