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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. 所以在遍历时要为每个节点标记是否访问过
For example,
Given board =
[ ["ABCE"], ["SFCS"], ["ADEE"] ]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
class Solution { public: bool exist(vector<vector<char> > &board, string word) { if(board.empty()) return false; vector<vector<bool>> visited(board.size(), vector<bool>(board[0].size(), false)); bool result = false; for(int i = 0; i < board.size(); i++) { for(int j = 0; j < board[0].size(); j++) { result = dfs(board,word,i,j,0,visited); if(result) return true; visited[i][j] = false; //回溯 } } return false; } bool dfs(vector<vector<char> > &board, string word, int i, int j, int depth, vector<vector<bool>> &visited) { if(board[i][j] != word[depth]) return false; if(depth == word.length()-1) return true; visited[i][j] = true; if(j < board[0].size()-1 && !visited[i][j+1]){ //向右 if(dfs(board,word, i, j+1, depth+1,visited)) return true; else visited[i][j+1] = false; //回溯 } if(j > 0 && !visited[i][j-1]) //向左 { if(dfs(board,word, i, j-1, depth+1,visited)) return true; else visited[i][j-1] = false; //回溯 } if(i < board.size()-1 && !visited[i+1][j]) //向下 { if(dfs(board,word, i+1, j, depth+1,visited)) return true; else visited[i+1][j] = false; //回溯 } if(i > 0 && !visited[i-1][j]) //向上 { if(dfs(board, word, i-1, j, depth+1,visited)) return true; else visited[i-1][j] = false; //回溯 } return false; } };
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/4854674.html
