Leetcode 79 Word Search

triplebee

发布于 2018-01-12 15:01:43

5520

发布于 2018-01-12 15:01:43

文章被收录于专栏：计算机视觉与深度学习基础计算机视觉与深度学习基础计算机视觉与深度学习基础

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.

在矩阵中找单词。

又是一个深搜，代码不够精简，写死了。

class Solution {
public:
    bool dfs(vector<vector<char>>& board, string word,int x,int y,vector<vector<bool>>& vis)
    {
        if(word.empty()) return true;
        if(x+1<board.size() && word[0]==board[x+1][y] && !vis[x+1][y])
        {
            vis[x+1][y]=true;
            if(dfs(board,word.substr(1,word.size()-1),x+1,y,vis)) return true;
            vis[x+1][y]=false;
        }
        if(x-1>=0 && word[0]==board[x-1][y] && !vis[x-1][y]) 
        {
            vis[x-1][y]=true;
            if(dfs(board,word.substr(1,word.size()-1),x-1,y,vis)) return true;
            vis[x-1][y]=false;
        }
        if(y+1<board[0].size() && word[0]==board[x][y+1] && !vis[x][y+1]) 
        {
            vis[x][y+1]=true;
            if(dfs(board,word.substr(1,word.size()-1),x,y+1,vis)) return true;
            vis[x][y+1]=false;
        }
        if(y-1>=0 && word[0]==board[x][y-1] && !vis[x][y-1]) 
        {
            vis[x][y-1]=true;
            if(dfs(board,word.substr(1,word.size()-1),x,y-1,vis)) return true;
            vis[x][y-1]=false;
        }
        return false;
    }
    bool exist(vector<vector<char>>& board, string word) {
        if(word.empty()) return true;
        vector<bool> temp(board[0].size(),false);
        vector<vector<bool>> vis(board.size(),temp);
        for(int i=0;i<board.size();i++)
            for(int j=0;j<board[0].size();j++)
                if(word[0]==board[i][j])
                {
                    vis[i][j]=true;
                    if(dfs(board,word.substr(1,word.size()-1),i,j,vis)) return true;
                    vis[i][j]=false;
                }
        
        return false;
    }
};

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原始发表：2016-09-23 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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