标签：strong sts int ons har 注意 example grid ace
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 =
[ [‘A‘,‘B‘,‘C‘,‘E‘], [‘S‘,‘F‘,‘C‘,‘S‘], [‘A‘,‘D‘,‘E‘,‘E‘] ]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
分析： 刚开始做的时候有误解，实际上就是从矩阵中找一条线路求解出word dfs求解最方便，注意一下边界范围
class Solution { private: int rows; int cols; public: bool exist(vector<vector<char>>& board, string word) { rows = board.size(); cols = board[0].size(); for(int i=0; i< rows; i++) for(int j=0;j<cols; j++){ if(dfs(board,i,j,word,0)) return true; } return false; } bool dfs(vector<vector<char>>& board, int x, int y,string word, int index){ //cout << x << " "<<y <<endl; //cout << "index = "<<index<<endl; if(x<0 || y<0 || x>=rows || y>=cols || (index<word.size() && board[x][y]!=word[index]) ) return false; if(index+1 == word.size()){ return true; } char c =board[x][y]; board[x][y] = ‘.‘; if(dfs(board,x-1,y,word,index+1) || dfs(board,x+1,y,word,index+1) || dfs(board,x,y-1,word,index+1) || dfs(board,x,y+1,word,index+1)) return true; board[x][y] =c; return false; } };
标签：strong sts int ons har 注意 example grid ace
原文地址：http://www.cnblogs.com/willwu/p/6250266.html
