79. Word Search

题目：

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,

Given board =

[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

word = 

"ABCCED"

,
-> returns 

true

,
word = 

"SEE"

,
-> returns 

true

,
word = 

"ABCB"

,
-> returns 

false

.
找字符列表中找字符串（dfs）

public class Solution {
public boolean exist(char[][] board, String word) {
int n= board.length;
int m= board[0].length;
int[][] map = new int[board.length][board[0].length];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(dfs(0,map,board,word,i,j))
{
return true;
}
}
}
return false;
}
int[][] dir ={{0,1},{0,-1},{1,0},{-1,0}};
private boolean dfs(int index,int[][] map,char[][] board,String word,int x,int y)
{

if(index == word.length())return true;
if(x<0 || x>=board.length || y<0 || y>=board[0].length)return false;
if(map[x][y]==1)return false;
if(board[x][y]!=word.charAt(index)) return false;
map[x][y]=1;
for(int i=0;i<4;i++)
{
int xx= x+dir[i][0];
int yy= y+dir[i][1];
if(dfs(index+1,map,board,word,xx,yy))return true;
}
map[x][y]=0;
return false;
}
}