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"]
]

又是关于搜索的一道题，简单利用深度优先搜索可以实现要求，用c++实现比较不容易超时，同样使用java写出来可能会出现超时 错误，语言的差异问题没办法。

class Solution {
private:
    vector<vector<char> > *board;
    string *word;
    bool **used;
private:
    bool isInboard(int i, int j)
    {
        if(i < 0)return false;
        if(i >= board->size())return false;
        if(j < 0)return false;
        if(j >= (*board)[i].size())return false;
        return true;
    }

    bool DFS(int si, int sj, int n)
    {
        if(n == word->size())return true;
        if(isInboard(si, sj))
      if(!used[si][sj] && (*board)[si][sj] == (*word)[n]){
            used[si][sj] = true;

            if(DFS(si+1, sj, n+1)){
                return true;
            }else if(DFS(si-1, sj, n+1)){
                return true;
            }else if(DFS(si, sj+1, n+1)){
                return true;
            }else if(DFS(si, sj-1, n+1)){
                return true;
            }

            //reset
            used[si][sj] = false;

            return false;
        }
        return false;
    }

public:
    bool exist(vector<vector<char> > &board, string word) {
        if(board.size() == 0)return false;
        this->board = &board;
        this->word = &word;
        used = new bool*[board.size()];
        for(int i = 0; i < board.size(); i ++)
        {
            used[i] = new bool[board[i].size()];
            for(int j = 0; j < board[i].size(); j ++)
                used[i][j] = false;
        }
        for(int i = 0; i < board.size(); i ++)
            for(int j = 0; j < board[i].size(); j ++)
                if(DFS(i, j, 0))return true;
        return false;
    }
};