79. Word Search

最新推荐文章于 2024-02-12 20:14:25 发布

原创最新推荐文章于 2024-02-12 20:14:25 发布 · 195 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

90 篇文章

订阅专栏

本文介绍了一个算法问题，即在一个二维字符网格中查找指定单词是否存在。通过深度优先搜索(DFS)递归地检查每个可能的路径，确保不重复访问同一单元格。文章提供了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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.

对于board中的任意一点如果是word中的第一个字母，则查找word中第二个字母是否在

board中该点四周，如果在，则递归至第二个字母，如此直至word的最后一个字母

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(word.size()==0) 
            return true;
        int row=board.size();
        int col=board[0].size();
        vector<vector<int>>visited(row, vector<int>(col)); //判断某个点是否遍历过
        if(row==0||col==0) return false;
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(board[i][j]==word[0])
                   if(dfs(board,word,0,i,j,visited))
                       return true;
            }
        }
        return false;
    }
    
    bool dfs(vector<vector<char>>&board,string word,int count,int row,int col,vector<vector<int>> &visited){
        if(count==word.size()-1)
            return true;
        visited[row][col]=1;  //访问该点设置已遍历
        if(row-1>=0&&visited[row-1][col]==0&&board[row-1][col]==word[count+1])//上下左右递归
            if(dfs(board,word,count+1,row-1,col,visited))
                return true;
        if(row+1<board.size()&&visited[row+1][col]==0&&board[row+1][col]==word[count+1])
            if(dfs(board,word,count+1,row+1,col,visited))
                return true;
        if(col-1>=0&&visited[row][col-1]==0&&board[row][col-1]==word[count+1])
            if(dfs(board,word,count+1,row,col-1,visited))
                return true;
        if(col+1<board[0].size()&&visited[row][col+1]==0&&board[row][col+1]==word[count+1])
            if(dfs(board,word,count+1,row,col+1,visited))
                return true;
        visited[row][col]=0;  //访问失败设置为0
        return false;
    }
};