79. 单词搜索 DFS回溯

79. 单词搜索

难度：中等
前几天小米笔试的真题来着，但是当时没给二维矩阵
题目描述

解题思路
回溯+剪枝+状态重置

 /*
					  * 79. 单词搜索
					  * 2020/9/14
					  * medium
					  */
					 public boolean exist(char[][] board, String word) {
						 
						 char[] str = word.toCharArray();
						 int m = board.length;
						 int n = board[0].length;
						 
						 for (int i = 0; i < m; i++) {
							for (int j = 0; j < n; j++) {
								if(board[i][j] == str[0]) {
									if(dfsExist(board,str,i,j,m,n,0))
										return true;
								}
							}
						}
						 
						 return false;

					 }
					 public boolean dfsExist(char[][] board,char[] str,int i,int j,int m,int n,int cur) {
						 
						 if(cur == str.length) {
							 return true;
						 }
						 if(i < 0 || i >= m || j < 0 || j >= n || board[i][j] == '#' || board[i][j] != str[cur]) {
							 return false;
						 }
						 char temp = board[i][j];
						 board[i][j] = '#';
						 boolean re = dfsExist(board,str,i+1,j,m,n,cur+1)
								 || dfsExist(board,str,i-1,j,m,n,cur+1)
								 || dfsExist(board,str,i,j+1,m,n,cur+1)
								 || dfsExist(board,str,i,j-1,m,n,cur+1);
						 
						 board[i][j] = temp;
						 return re;
						 
					 }