Leetcode79 Word Search

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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/word-search

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

思路就是回溯法。中间纠结了一下，visited的数组是用二维数组还是用vector。后来发现如果用二维数组，传指针的地方较为麻烦，而且不能像vector一样直接visited[row][col]。