2014-05-01 02:30

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
     
      
        > &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
       
        
          > &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 here48             // letter mismatch here49             return false;50         }51         52         bool res;53         if (idx == word_len - 1) {54             // reach the end of word, success55             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 };