Word Search

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"]
]
解题思路： 类似于迷宫，递归回溯。需要一个辅助数组记录走过的位置，防止同一个位置被使用多次。

public class Solution {
    public boolean exist(char[][] board, String word) {
        if(word==null){
            return true;
        }
        boolean bool[][]=new boolean[board.length][board[0].length];
        for(int i=0;i<board.length;i++){
            for(int j=0;j<board[i].length;j++){
                bool[i][j]=false;
            }
        }
         for(int i=0;i<board.length;i++){
            for(int j=0;j<board[i].length;j++){
                if(board[i][j]==word.charAt(0)){
                    bool[i][j]=true;
                  if(search(board,word.substring(1),i,j,bool))//注意：当search返回false时，还要继续遍历，记得把bool置为false
                    return true;
                  else
                    bool[i][j]=false;
                }
                
            }
        }
        return false;
        
    }
    boolean search(char[][] board,String word,int i,int j,boolean[][]bool){
        if(word.length()==0) return true;//判断条件根据length，不是null
        int [][]director={{-1,0},{1,0},{0,-1},{0,1}};//此处非常有技巧，代表四个方向
        for(int k=0;k<director.length;k++){
            int ii=i+director[k][0];
            int jj=j+director[k][1];
            if(ii>=0&&ii<board.length&&jj>=0&&jj<board[0].length){
                 if(board[ii][jj]==word.charAt(0)&&bool[ii][jj]==false){
                    bool[ii][jj]=true;
                 if(search(board,word.substring(1),ii,jj,bool))
                    return true;
                 else 
                    bool[ii][jj]=false;
            }
                
            }
           
        }
        return false;
        
        
        
    }
}