标签：
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.
class Solution { private: int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; public: bool help(vector<vector<char>> &board,vector<vector<bool>> &pos,string &word,int cur,int x,int y){ if(cur==word.length()) return true; for(int i=0;i<4;i++){ int pos_x=x+dir[i][0]; int pos_y=y+dir[i][1]; if(pos_x>=0&&pos_x<board.size()&&pos_y>=0&&pos_y<board[0].size()&&!pos[pos_x][pos_y]&&board[pos_x][pos_y]==word[cur]){ pos[pos_x][pos_y]=true; if(help(board,pos,word,cur+1,pos_x,pos_y)) return true; pos[pos_x][pos_y]=false; } } return false; } bool exist(vector<vector<char>>& board, string word) { if(word.length()==0) return true; int m=board.size(); int n=board[0].size(); vector<vector<bool>>pos(m,vector<bool>(n,false)); for(int i=0;i<m;i++){ for(int j=0;j<n;j++){ if(word[0]==board[i][j]){ pos[i][j]=true; if(help(board,pos,word,1,i,j)) return true; pos[i][j]=false; } } } return false; } };
标签：
原文地址：http://www.cnblogs.com/zhoudayang/p/5071716.html
