## Monday, January 14, 2013

### [LeetCode] Word Search 解题报告

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"],
]
```
word = `"ABCCED"`, -> returns `true`,
word = `"SEE"`, -> returns `true`,
word = `"ABCB"`, -> returns `false`.
» Solve this problem

[解题思路]

[Code]
``````1:    bool exist(vector<vector<char> > &board, string word) {
2:      // Start typing your C/C++ solution below
3:      // DO NOT write int main() function
4:      if(word.size() ==0) return false;
5:      if(board.size() ==0 || board[0].size() == 0) return false;
6:      int row = board.size();
7:      int col = board[0].size();
8:      int * visited = new int[row*col];
9:      memset(visited, 0, row*col*sizeof(int));
10:      for(int i =0; i< board.size(); i++)
11:      {
12:        for(int j =0; j< board[0].size(); j++)
13:        {
14:          if(board[i][j] == word[0])
15:          {
16:            visited[i*col+j] = 1;
17:            if(search(board, word, visited, -1, 1, i, j))
18:              return true;
19:            visited[i*col+j] =0;
20:          }
21:        }
22:      }
23:      delete visited;
24:      return false;
25:    }
26:    bool search(vector<vector<char> > &board,
27:      string& word,
28:      int* visited,
29:      int op, //0 up, 1 down, 2 left, 3 right
30:      int matchLen,
31:      int i,
32:      int j)
33:    {
34:      if(matchLen == word.size()) return true;
35:      int row = board.size();
36:      int col = board[0].size();
37:      if(i+1<row && op!=0)
38:      {
39:        if(visited[(i+1)*col+j] ==0 &&
40:        board[i+1][j] == word[matchLen])
41:        {
42:          visited[(i+1)*col+j] =1;
43:          if(search(board, word, visited, 1, matchLen+1, i+1, j))
44:            return true;
45:          visited[(i+1)*col+j] =0;
46:        }
47:      }
48:      if(i-1>=0 && op!=1)
49:      {
50:        if(visited[(i-1)*col+j] ==0 && board[i-1][j] == word[matchLen])
51:        {
52:          visited[(i-1)*col+j] =1;
53:          if(search(board, word, visited, 0, matchLen+1, i-1, j))
54:            return true;
55:          visited[(i-1)*col+j] =0;
56:        }
57:      }
58:      if(j+1<col && op!=2)
59:      {
60:        if(visited[i*col+j+1] ==0 && board[i][j+1] == word[matchLen])
61:        {
62:          visited[i*col+j+1] =1;
63:          if(search(board, word, visited, 3, matchLen+1, i, j+1))
64:            return true;
65:          visited[i*col+j+1] =0;
66:        }
67:      }
68:      if(j-1>=0 && op!=3)
69:      {
70:        if(visited[i*col+j-1] ==0 && board[i][j-1] == word[matchLen])
71:        {
72:          visited[i*col+j-1] =1;
73:          if(search(board, word, visited, 2, matchLen+1, i, j-1))
74:            return true;
75:          visited[i*col+j-1] =0;
76:        }
77:      }
78:      return false;
79:    }
``````