# 题目描述

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.

Example:

board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

# 代码实现

class Solution {
public:
bool existCore(vector<vector<char>>& board, string &word,int & Wordpos,int rows,int cols,int row,int col,vector<vector<int> > & visited){
if(Wordpos>=word.size())
return true;
if(col>=cols||row>=rows||col<0||row<0){
return false;
}
bool flag=false;
if(board[row][col]==word[Wordpos]&& visited[row][col]!=1){
++Wordpos;
visited[row][col]=1;
flag=existCore(board,word,Wordpos,rows,cols,row+1,col,visited)||existCore(board,word,Wordpos,rows,cols,row,col+1,visited)||existCore(board,word,Wordpos,rows,cols,row-1,col,visited)||existCore(board,word,Wordpos,rows,cols,row,col-1,visited);
if(!flag)
{
--Wordpos;
visited[row][col]=0;
}
}
return flag;
}
bool exist(vector<vector<char>>& board, string word) {
bool result=false;
if(board.size()==0||board[0].size()==0){
return false;
}
int rows=board.size();
int cols=board[0].size();
vector<vector<int> > visited;
for(int i=0;i<rows;++i){
vector<int> temp;
for(int j=0;j<cols;++j){
temp.push_back(0);
}
visited.push_back(temp);
}
int Wordpos=0;
for(int i=0;i<rows;++i){
for(int j=0;j<cols;++j)
{
if(existCore(board,word,Wordpos,rows,cols,i,j,visited))
return true;
}
}
return false;
}
};


# 总结

posted @ 2019-09-25 20:47  HaoPeng_Zhang  阅读(77)  评论(0编辑  收藏