[LeetCode] Word Search

This problem is somewhat tricky at first glance. However, the final implementation is fairly simple using recursion.

The basic idea is, visiting every possible position (i, j) of  board and find if starting from (i, j), the word exists in the board. For the word to exist starting from (i, j), two conditions should be satisfied:

  1. board[i][j] = word[0] ;
  2. Define tail = word.substr(1, word.length() - 1) , then we should be able to find tail starting from one of the four neighbors of (i, j), which requires a recursive call of the function.

Putting 1 and 2 together, we have the following code, which is self-explanatory.

 1 class Solution {
 2 public:
 3     bool exist(vector<vector<char>> board, string word) {
 4         for (int i = 0; i < board.size(); i++)
 5             for (int j = 0; j < board[0].size(); j++)
 6                 if (isExist(board, word, i, j)) return true;
 7         return false;
 8     }
 9 private:
10     bool isExist(vector<vector<char>>& board, string& word, int row, int col) {
11         if (row < 0 || row >= board.size() || col < 0 || col >= board[0].size() || board[row][col] != word[0])
12             return false;
13         if (word.length() == 1) return true;
14         board[row][col] = ' ';
15         string tail = word.substr(1, word.length() - 1);
16         if (isExist(board, tail, row - 1, col) || isExist(board, tail, row + 1, col) ||
17             isExist(board, tail, row, col - 1) || isExist(board, tail, row, col + 1))
18             return true;
19         board[row][col] = word[0];
20         return false;
21     }
22 };

Note that in the above code, in order to prevent visiting the same grid of board again, we modify the starting point of the grid and recover it again in lines 14 and 19 respectively.

The above code is readable, though not fast enough. A simple trick to make it faster to pass a char*  instead of string to the isExist function.

 1 class Solution {
 2 public:
 3     bool exist(vector<vector<char>> board, string word) {
 4         for (int i = 0; i < board.size(); i++)
 5             for (int j = 0; j < board[0].size(); j++)
 6                 if (isExist(board, word.c_str(), i, j)) return true;
 7         return false;
 8     }
 9 private:
10     bool isExist(vector<vector<char>>& board, const char* word, int row, int col) {
11         if (row < 0 || row >= board.size() || col < 0 || col >= board[0].size() || board[row][col] != *word)
12             return false;
13         if (!(*(word + 1))) return true;
14         board[row][col] = ' ';
15         if (isExist(board, word + 1, row - 1, col) || isExist(board, word + 1, row + 1, col) ||
16             isExist(board, word + 1, row, col - 1) || isExist(board, word + 1, row, col + 1))
17             return true;
18         board[row][col] = *word;
19         return false;
20     }
21 };

 

posted @ 2015-06-04 10:00  jianchao-li  阅读(257)  评论(0)  编辑  收藏  举报