leetcode79. 单词搜索 网格地图搜索+回溯经典写法啦

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.

思路：搜索回溯基本上是经典模板题了。

class Solution {
    private boolean[][] marked;
    //        x-1,y
    // x,y-1  x,y    x,y+1
    //        x+1,y
    private int[][] direction = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
    // 盘面上有多少行
    private int m;
    // 盘面上有多少列
    private int n;
    private String word;
    private char[][] board;
    public boolean exist(char[][] board, String word) {
        m = board.length;
        if (m == 0)return false;
        n = board[0].length;
        marked = new boolean[m][n];
        this.word = word;
        this.board = board;

        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (dfs(i, j, 0))
                    return true;
        return false;
    }

    private boolean dfs(int i, int j, int start) {
        if (start == word.length() - 1) {
            return board[i][j] == word.charAt(start);
        }
        if (board[i][j] == word.charAt(start)) {
            marked[i][j] = true;
            for (int k = 0; k < 4; k++) {
                int newX = i + direction[k][0];
                int newY = j + direction[k][1];
                if (newX >= 0 && newX < m && newY >= 0 && newY < n && !marked[newX][newY]) {
                    if (dfs(newX, newY, start + 1)) {
                        return true;
                    }
                }
            }
            marked[i][j] = false;
        }
        return false;
    }
}