Word Search [LeetCode]

Problem Description: http://oj.leetcode.com/problems/word-search/

Basic idea: recursively go forward, use char '#' to mark the char visited, so no extra memory is need, don't forget to recover the previous char if the program cannot go forward.

 class Solution {
 public:
     bool existSub(int i, int j, vector<vector<char> > &board, string word){
         if(word.size() == )
             return true;
         if(i <  || j <  || i >= board.size() || j >= board[].size())
             return false;
         if(board[i][j] != word[])
             return false;

         string sub_word = word.substr();
         char ch = board[i][j];
         board[i][j] = '#';
         bool ret = existSub(i - , j,  board, sub_word) ||
                 existSub(i + , j, board, sub_word) ||
                 existSub(i, j - , board, sub_word) ||
                 existSub(i, j + , board, sub_word);
         if (ret)
             return true;

         board[i][j] = ch;
         return false;
     }

     bool exist(vector<vector<char> > &board, string word) {
         // Note: The Solution object is instantiated only once and is reused by each test case.
         if(word.size() == )
             return true;

         for(int i = ; i < board.size(); i++)
             for(int j = ; j < board[i].size(); j++)
                 if(board[i][j] == word[])
                     if(existSub(i, j, board, word))
                         return true;

         return false;
     }
 };