Leetcode-79. Word Search

最新推荐文章于 2023-06-03 09:46:57 发布

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原文链接：http://www.cnblogs.com/hongyang/p/6061704.html

本文介绍了一个简单的WordSearch算法实现，该算法能够在二维网格中查找指定单词。通过深度优先搜索(DFS)的方式，从每个可能的位置开始尝试匹配单词，并确保同一字母单元格不会被重复使用。文章提供了完整的C++代码实现及解析。

79. Word Search

QuestionEditorial Solution

My Submissions

Total Accepted: 98890
Total Submissions: 397019
Difficulty: Medium
Contributors: Admin

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.

思路：

四个方向挨个搜索即可，简单搜索题。

code:

 1 class Solution {
 2 public:
 3     int row,col;
 4     bool exist(vector<vector<char>>& board, string word) {
 5         row=board.size();
 6         if(row==0)
 7         return false;
 8         col=board[0].size();
 9 
10         vector<vector<int>> flag(row, vector<int>(col));
11         for(int i=0;i<board.size();i++)
12         {
13             for(int j=0;j<board[0].size();j++)
14             {
15                 //memset(flag,0,sizeof(flag));
16                 if(dfs(board,word,word.length(),i,j,flag))
17                 return true;
18             }
19         }
20         
21         return false;
22     }
23     
24     bool dfs(vector<vector<char>>& map,string word,int sum,int x,int y,vector<vector<int>>& flag)
25     {
26         if(sum==0)
27         return true;
28         bool res=false;
29         if(legal(x,y)&&map[x][y]==word[word.length()-sum]&&flag[x][y]!=1)
30         {
31             flag[x][y]=1;
32             res=res||dfs(map,word,sum-1,x+1,y,flag);
33             res=res||dfs(map,word,sum-1,x-1,y,flag);
34             res=res||dfs(map,word,sum-1,x,y-1,flag);
35             res=res||dfs(map,word,sum-1,x,y+1,flag);
36             flag[x][y]=0;
37         }
38         return res;
39     }
40     
41     bool legal(int x,int y)
42     {
43         if(x>=0&&x<row&&y>=0&&y<col)
44         return true;
45         return false;
46     }
47 };