使用的别人的思路，用一个二维数组记录每一个位置是否用过，然后通过递归来判断每一个位置是否符合
public class Solution {
public boolean exist(char[][] board, String word) {
if(board.length == 0) return false;
int leni = board.length;
int lenj = board[0].length;
boolean[][] isVisited = new boolean[leni][lenj];
for(int i=0;i < leni;++i){
for(int j = 0; j < lenj;++j){
isVisited[i][j]= false;
}
}
for(int i = 0; i < leni;++i){
for(int j = 0 ; j < lenj;++j){
if(board[i][j] == word.charAt(0))
{
isVisited[i][j] = true;
if(WordSearch(board,isVisited,word.substring(1),i,j)){
return true;
}
isVisited[i][j] = false;
}
}
}
return false;
}
public boolean WordSearch(char[][] board,boolean[][] isVisited,String word,int i, int j){
if(word.length() == 0) return true;
int[][] direction={{0,1},{0,-1},{-1,0},{1,0}};//上下左右
for(int k = 0; k < direction.length;++k){
int x = i + direction[k][0];
int y = j + direction[k][1];
if((x >= 0 && x < board.length) &&
(y >= 0 && y < board[i].length) &&
board[x][y] == word.charAt(0)&&
isVisited[x][y] == false){
isVisited[x][y] = true;
if(WordSearch(board,isVisited,word.substring(1), x, y)){
return true;
}
isVisited[x][y] = false;
}
}
return false;
}
}