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 =
[ ["ABCE"], ["SFCS"], ["ADEE"] ]word =
"ABCCED", -> returns true,"SEE", -> returns true,"ABCB", -> returns false.
public class Solution {
boolean [][]visit;
char[][]board;
int ROW;
int COL;
public boolean exist(int row,int col,String word)
{
if(board[row][col]!=word.charAt(0))
{
return false;
}
if(word.length()==1)
{
return true;
}
boolean result=false;
visit[row][col]=true;
String remaining=word.substring(1);
if(row>0&&!visit[row-1][col])
{
result=exist(row-1,col,remaining);
}
if(!result&&row<ROW-1&&!visit[row+1][col])
{
result=exist(row+1,col,remaining);
}
if(!result&&col>0&&!visit[row][col-1])
{
result=exist(row,col-1,remaining);
}
if(!result&&col<COL-1&&!visit[row][col+1])
{
result=exist(row,col+1,remaining);
}
visit[row][col]=false;
return result;
}
public boolean exist(char[][] board, String word) {
if(word==null||word.length()==0)
{
return true;
}
if(board==null)
{
return false;
}
this.board=board;
ROW=board.length;
COL=board[0].length;
visit=new boolean[ROW][COL];
for(int i=0;i<ROW;i++)
{
for(int j=0;j<COL;j++)
{
visit[i][j]=false;
}
}
for(int i=0;i<ROW;i++)
{
for(int j=0;j<COL;j++)
{
if(exist(i,j,word)){
return true;
}
}
}
return false;
}
}原文地址：http://blog.csdn.net/jiewuyou/article/details/39348849
