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.

思路：backtracking经典题目，从一个点开始，搜索所有可能的解空间，然后还原这个点，然后从下一个点开始。

注意，用for循环来check每一个点，然后以每一个点开始搜索，这样可以快很多，不需要搜索所有的解空间，找到合适的解，即刻返回true就可以了。

一个小技巧是因为board是char array，所以只用判断word.charAt(index)就可以了，不需要用stringbuilder。

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