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：

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

示例 2：

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

示例 3：

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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

进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？

棋盘格的DFS需要遍历整个棋盘格，对每个格子进行深搜。

dfs判断能不能从board[i][j]搜到word.charAt(index)

如果能搜到就搜下一个index,如果index到达最后一个位置说明前边都符合，返回true

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class Solution {
    int m = 0;
    int n = 0;

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

    // 返回值是board[i][j]能否匹配到word[index]之后的所有
    private boolean dfs(char[][] board, String word, int i, int j, int index) {
        if (board[i][j] != word.charAt(index)) {
            return false;
        }
        if (index == word.length() - 1) {
            return true;
        }
        int dx[] = { 0, 1, 0, -1 };
        int dy[] = { 1, 0, -1, 0 };
        board[i][j] = 0;
        for (int d = 0; d < 4; d++) {
            int x = i + dx[d];
            int y = j + dy[d];
            if (x >= 0 && x < m && y >= 0 && y < n && dfs(board, word, x, y, index + 1)) {
                return true;
            }
        }
        board[i][j] = word.charAt(index);
        return false;
    }
}