Careercup - Facebook面试题 - 5890898499993600

2014-05-01 02:30

题目链接

原题：

Given a matrix of letters and a word, check if the word is present in the matrix. E,g., suppose matrix is: 

a b c d e f 

z n a b c f 

f g f a b c 

and given word is fnz, it is present. However, gng is not since you would be repeating g twice. 

You can move in all the 8 directions around an element.

题目：给定一个字母矩阵，给定一个单词，请判断从矩阵某一点出发，能否走出一条路径组成这个单词。每一步可以走8邻接的方向。

解法：这基本就是Leetcode的原题Word Search，DFS搞定。如果矩阵太大的话，可以用BFS防止递归过深造成的栈溢出，不过效率方面就更慢了。

代码：

 1 // http://www.careercup.com/question?id=5890898499993600

 2 class Solution {

 3 public:

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

 5         n = (int)board.size();

 6         if (n == 0) {

 7             return false;

 8         }

 9         m = (int)board[0].size();

10         word_len = (int)word.length();

11         

12         if (word_len == 0) {

13             return true;

14         }

15         

16         int i, j;

17         for (i = 0; i < n; ++i) {

18             for (j = 0; j < m; ++j) {

19                 if(dfs(board, word, i, j, 0)) {

20                     return true;

21                 }

22             }

23         }

24         return false;

25     }

26 private:

27     int n, m;

28     int word_len;

29     

30     bool dfs(vector<vector<char> > &board, string &word, int x, int y, int idx) {

31         static const int d[8][2] = {

32             {-1, -1}, 

33             {-1,  0}, 

34             {-1, +1}, 

35             { 0, -1}, 

36             { 0, +1}, 

37             {+1, -1}, 

38             {+1,  0}, 

39             {+1, +1}

40         };

41         

42         if (x < 0 || x > n - 1 || y < 0 || y > m - 1) {

43             return false;

44         }

45         

46         if (board[x][y] < 'A' || board[x][y] != word[idx]) {

47             // already searched here

48             // letter mismatch here

49             return false;

50         }

51         

52         bool res;

53         if (idx == word_len - 1) {

54             // reach the end of word, success

55             return true;

56         }

57 

58         for (int i = 0; i < 8; ++i) {

59             board[x][y] -= 'a';

60             res = dfs(board, word, x + d[i][0], y + d[i][1], idx + 1);

61             board[x][y] += 'a';

62             if (res) {

63                 return true;

64             }

65         }

66         // all letters will be within [a-z], thus I marked a position as 'searched' by setting them to an invalid value.

67         // we have to restore the value when the DFS is done, so their values must still be distiguishable.

68         // therefore, I used an offset value of 'a'.

69         // this tricky way is to save the extra O(n * m) space needed as marker array.

70         

71         return false;

72     }

73 };