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.
分析：dfs。
1 class Solution { 2 public: 3 bool exist(vector<vector<char> > &board, string word) { 4 int n = board.size(); 5 if(n == 0) return false; 6 int m = board[0].size(); 7 if(m == 0) return false; 8 vector<vector<bool> > used(n, vector<bool>(m,false)); 9 10 for(int i = 0; i < n; i++) 11 for(int j = 0; j < m; j++){ 12 if(dfs(board,word,used,i,j,0)) 13 return true; 14 } 15 return false; 16 } 17 bool dfs(vector<vector<char> > &board, string word, vector<vector<bool> > &used, int i, int j, int k){ 18 if(k == word.length()) return true; 19 if(i < 0 || j < 0 || i >= board.size() || j >= board[0].size()) return false;//first check range 20 if(board[i][j] != word[k] || used[i][j]) return false;//then check this 21 used[i][j] = true; 22 if(dfs(board, word,used,i-1,j,k+1) || dfs(board, word, used, i+1,j,k+1) || dfs(board, word, used, i, j-1, k+1) 23 || dfs(board, word, used, i, j+1, k+1)) 24 return true; 25 used[i][j] = false; 26 return false; 27 28 } 29 };
标签：style blog io ar color os sp for on
原文地址：http://www.cnblogs.com/Kai-Xing/p/4149038.html
