leetcode 79. 单词搜索

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

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 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true
示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true
示例 3：

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

提示：

m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board 和 word 仅由大小写英文字母组成

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/word-search
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回溯一个一个的找，没有申请额外的空间来记录路程，因为都是字符，所以用数字 0 来标记路程。

    public boolean exist(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                boolean b = find(board, word, i, j, 0);
                if (b) {
                    return b;
                }
            }
        }
        return false;
    }

    private static boolean find (char[][] board, String word, int x, int y, int index) {
        int m = board.length;
        int n = board[0].length;
        if (x < 0 || x == m) {
            return false;
        }
        if (y < 0 || y == n) {
            return false;
        }
        char c = board[x][y];
        if (c == '0') {
            return false;
        }
        boolean flag = false;
        if (c == word.charAt(index)) {
            index++;
            if (word.length() == index) {
                return true;
            } 
            board[x][y] = '0';
            flag = find(board, word, x, y + 1, index) 
                    || find(board, word, x, y - 1, index) 
                    || find(board, word, x + 1, y, index) 
                    || find(board, word, x - 1, y, index);
            board[x][y] = c;
        }
        return flag;
    }