Word Search——经典题（还没细看）-优快云博客

本文介绍了一个二维网格中查找单词的问题及解决方案。通过回溯算法，文章详细解释了如何判断一个单词是否能由网格中相邻字母构成，且同一字母单元格不能重复使用。代码示例清晰展示了如何利用递归和标记已访问单元格的方法来实现这一过程。

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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 =

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

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

这也算是设置标志位的经典例子了吧。(回溯)

class Solution {
public:
    bool isExist(vector<vector<char>>& board,string wordsub,int h,int w, vector<vector<bool>>&flag)
    {
        int direct[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
        for(int d=0;d<4;d++)
        {
            int jj=h+direct[d][0];
            int ii=w+direct[d][1];
            if(jj>=0&&jj<board.size()&&ii>=0&&ii<board[0].size())
            {
                if(wordsub[0]==board[jj][ii]&&flag[jj][ii]==0)
                {
                    flag[jj][ii]=1;
                    if(wordsub.size()==1||isExist(board,wordsub.substr(1),jj,ii,flag))
                        return true;
                    flag[jj][ii]=0;
                }
                
            }
                
        }
        return false;
    }
   
    bool exist(vector<vector<char>>& board, string word) {
        int length=word.size();
        int Height=board.size();
        int Width=board[0].size();
        vector<vector<bool>> flag(Height,vector<bool>(Width,0));
        for(int h=0;h<Height;h++)
            for(int w=0;w<Width;w++)
            {
                if(board[h][w]==word[0])
                {
                    flag[h][w]=1;
                    if( word.size()==1||isExist(board,word.substr(1),h,w,flag))
                        return true;
                    flag[h][w]=0;
                }
            }
        return false;
    }
};