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.
题解：
public class L79 { static Boolean resflag = false; static int[] dx = {1,0,-1,0},dy = {0,-1,0,1}; public static boolean exist(char[][] board, String word) { resflag = false; if (board == null || board.length == 0){return false;} char[] wordChar = word.toCharArray(); boolean[][] flag = new boolean[board.length][board[0].length]; for(int boardX = 0;boardX<board.length;boardX++){ for(int boardY = 0;boardY < board[0].length;boardY ++){ if(board[boardX][boardY] == wordChar[0]){ flag[boardX][boardY] = true; backExist(board,1,wordChar, flag,boardX,boardY); //结果为true，直接返回，否则继续遍历 if(resflag)return true; flag[boardX][boardY] = false;
} } } return false; } private static void backExist(char[][] board, int index, char[] wordChar, boolean[][] flag, int x, int y) { if(resflag || index == wordChar.length){ resflag = true; return; } //该步骤的判断是为了进行剪枝的操作，即目标完成后进行返回 for(int i=0;i<4;i++){ int X_next = x + dx[i];int Y_next = y + dy[i]; //不可以超出边界 if(X_next>=0 && Y_next>=0 & X_next< board.length & Y_next< board[0].length){ if (!(flag[X_next][Y_next]) && board[X_next][Y_next] == wordChar[index]){ flag[X_next][Y_next] = true; backExist(board,index+1,wordChar, flag, X_next,Y_next); flag[X_next][Y_next] = false; } } } return; } public static void main(String[] args) { char[][] board = {{'A','B','C','E'},{'S','F','E','S'},{'A','D','E','E'}}; String word = "ABCESEEEFS"; Boolean x = exist(board, word); System.out.println(x); }}