Word Search, wordsearch

Word Search, wordsearch

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,
GivenBoard=

[  ["ABCE"],  ["SFCS"],  ["ADEE"]]

Word= "ABCCED",-> Returns true,
Word= "SEE",-> Returns true,
Word= "ABCB",-> Returns false.

It is also a question about search. Simply using deep-priority search can meet the requirements. Using c ++ is not easy to implement timeout, and a timeout error may occur when writing in java, there is no way to solve the problem of language differences.

 

class Solution {private:    vector<vector<char> > *board;    string *word;    bool **used;private:    bool isInboard(int i, int j)    {        if(i < 0)return false;        if(i >= board->size())return false;        if(j < 0)return false;        if(j >= (*board)[i].size())return false;        return true;    }    bool DFS(int si, int sj, int n)    {        if(n == word->size())return true;        if(isInboard(si, sj))      if(!used[si][sj] && (*board)[si][sj] == (*word)[n]){            used[si][sj] = true;            if(DFS(si+1, sj, n+1)){                return true;            }else if(DFS(si-1, sj, n+1)){                return true;            }else if(DFS(si, sj+1, n+1)){                return true;            }else if(DFS(si, sj-1, n+1)){                return true;            }            //reset            used[si][sj] = false;            return false;        }        return false;    }public:    bool exist(vector<vector<char> > &board, string word) {        if(board.size() == 0)return false;        this->board = &board;        this->word = &word;        used = new bool*[board.size()];        for(int i = 0; i < board.size(); i ++)        {            used[i] = new bool[board[i].size()];            for(int j = 0; j < board[i].size(); j ++)                used[i][j] = false;        }        for(int i = 0; i < board.size(); i ++)            for(int j = 0; j < board[i].size(); j ++)                if(DFS(i, j, 0))return true;        return false;    }};

Word search

You can click Insert, image, clip art, and media repository to search for the cropped image. Enter the animal name. If you do not have this image, you can search for the image on the Internet using WORD Microsoft Office Online.

What is Word search?

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