标签：style blog io ar color os sp for on
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"], ["ADEE"] ]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
class Solution { private: vector<vector<char> > *board; string *word; bool **used; private: bool isInboard(int i, int j) { if(i < 0)return false; if(i >= board->size())return false; if(j < 0)return false; if(j >= (*board)[i].size())return false; return true; } bool DFS(int si, int sj, int n) { if(n == word->size())return true; if(isInboard(si, sj)) { if(!used[si][sj] && (*board)[si][sj] == (*word)[n]) { used[si][sj] = true; bool ret = false; if(DFS(si+1, sj, n+1)) ret = true; else if(DFS(si-1, sj, n+1)) ret = true; else if(DFS(si, sj+1, n+1)) ret = true; else if(DFS(si, sj-1, n+1)) ret = true; used[si][sj] = false; return ret; } } return false; } public: bool exist(vector<vector<char> > &board, string word) { if(board.size() == 0)return false; this->board = &board; this->word = &word; used = new bool*[board.size()]; for(int i = 0; i < board.size(); i ++) { used[i] = new bool[board[i].size()]; for(int j = 0; j < board[i].size(); j ++) used[i][j] = false; } for(int i = 0; i < board.size(); i ++) for(int j = 0; j < board[i].size(); j ++) if(DFS(i, j, 0))return true; return false; } };
标签：style blog io ar color os sp for on
原文地址：http://www.cnblogs.com/code-swan/p/4141215.html
