zoukankan      html  css  js  c++  java
  • Leetcode-79. Word Search

    79. Word Search

     
     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 };
  • 相关阅读:
    [NOI2014]动物园 题解(预览)
    CF1200E 题解
    KMP算法略解
    [EER2]谔运算 口胡
    CF504E Misha and LCP on Tree 题解
    长链剖分 解 k级祖先问题
    双哈希模板
    Luogu P5333 [JSOI2019]神经网络
    UOJ449 【集训队作业2018】喂鸽子
    LOJ6503 「雅礼集训 2018 Day4」Magic
  • 原文地址:https://www.cnblogs.com/hongyang/p/6061704.html
Copyright © 2011-2022 走看看