回溯法

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.

解答：

public class Solution {

    public static boolean exist(char[][] board, String word) {
        if(word.length() > board.length*board[0].length()) {
            return false;
        }

        boolean[][] boardFlag = new boolean[board.length][board[0].length];
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                if(find(i, j, board, word, boardFlag)) {
                    return true;
                }
            }
        }

        return false;
    }

    public static boolean find(int i, int j, char[][] board, String word, boolean[][] boardFlag) {
        if(board[i][j] == word.charAt(0) && !boardFlag[i][j]) {

            boardFlag[i][j] = true;

            if(word.length-1 == 0) {
                return true;
            }

            if(j + 1 < board[0].length) {
                if(find(i,j+1,board,word.substring(1,word.length()),boardFlag)) return true;
            }

            if(j - 1 >= 0) {
                if(find(i,j-1,board,word.substring(1,word.length()),boardFlag)) return true;
            }

            if (i+1 < board.length) {
                if(find(i+1,j,board,word.substring(1,word.length()),boardFlag)) return true;  
            }
                
            if (i-1 >= 0) {
                if(find(i-1,j,board,word.substring(1,word.length()),boardFlag)) return true;  
            }

            // important here
            boardFlag[i][j] = false;

            return false;
        }
    }
}