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.

 1 class Solution {

 2 public:

 3     bool existSub(int i, int j, vector<vector<char> > &board, string word){

 4         if(word.size() == 0)

 5             return true;

 6         if(i < 0 || j < 0 || i >= board.size() || j >= board[0].size())

 7             return false;

 9         if(board[i][j] != word[0])

10             return false;

11         

12         string sub_word = word.substr(1);

14         char ch = board[i][j];

15         board[i][j] = '#';

16         bool ret = existSub(i - 1, j,  board, sub_word) || 

17                 existSub(i + 1, j, board, sub_word) || 

18                 existSub(i, j - 1, board, sub_word) || 

19                 existSub(i, j + 1, board, sub_word);            

21         if (ret) 

22             return true;

23             

24         board[i][j] = ch;

25         return false;

26     }

27     

28     bool exist(vector<vector<char> > &board, string word) {

29         // Note: The Solution object is instantiated only once and is reused by each test case.

30         if(word.size() == 0)

31             return true;

32             

33         for(int i = 0; i < board.size(); i++)

34             for(int j = 0; j < board[i].size(); j++) 

35                 if(board[i][j] == word[0])

36                     if(existSub(i, j, board, word))

37                         return true;

38 

39         return false;

40     }

41 };