Leetcode题目79.单词搜索（回溯+DFS-中等）

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.

代码实现：

class Solution {
     public static boolean exist(char[][] board, String word) {

        if (board.length == 0) {
            return false;
        }
        //网格的行数
        int row = board.length;
        //网格的列数
        int col = board[0].length;
        //表示网格是否已经被访问过的状态
        boolean[][] visited = new boolean[row][col];
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (board[i][j] == word.charAt(0) && backTrace(i, j, 0, board, word, visited)) {
                    return true;
                }
            }
        }
        return false;
    }

    //DFS+回溯
    private static boolean backTrace(int i, int j, int index, char[][] board, String word, boolean[][] visited) {

        if (index == word.length()) {
            return true;
        }
        //二维平面按照上->右->下->左的顺序搜索
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != word.charAt(index) || visited[i][j]) {
            return false;
        }
        //标记当前网格已经被访问
        visited[i][j] = true;
        if (backTrace(i - 1, j, index + 1, board, word, visited)
                || backTrace(i, j + 1, index + 1, board, word, visited)
                || backTrace(i + 1, j, index + 1, board, word, visited)
                || backTrace(i, j - 1, index + 1, board, word, visited)) {

            return true;
        }
        //回溯
        visited[i][j] = false;
        return false;
    }
}