79. 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 =

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

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

本题重点在于体会回溯的方法，需要注意的是每次访问的时候需要加一个boolean值来确定是否访问过，代码如下：

public class Solution {
    public boolean exist(char[][] board, String word) {
        if(board==null||board.length==0||board[0].length==0) return false;
        if(word.length()==0) return true;
        int m = board.length;
        int n= board[0].length;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                boolean[][] tested = new boolean[m][n];
                if(helper(board,word,tested,i,j,0)){
                    return true;
                }
            }
        }
        return false;
    }
    public boolean helper(char[][] board,String word,boolean[][] tested,int row,int col,int cur){
        if(cur==word.length()) return true;
        if(row<0||row>=board.length||col<0||col>=board[0].length||board[row][col]!=word.charAt(cur)||tested[row][col]==true) return false;
        tested[row][col] = true;
        boolean exist=helper(board,word,tested,row+1,col,cur+1)||helper(board,word,tested,row,col+1,cur+1)||helper(board,word,tested,row-1,col,cur+1)||helper(board,word,tested,row,col-1,cur+1);
        tested[row][col]=false;
        return exist;
    }
}

 本题可以不用创建boolean数组，只要让那个位置的值异或256。