• Valid Word Square


    Given a sequence of words, check whether it forms a valid word square.

    A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns).

    Note:

    1. The number of words given is at least 1 and does not exceed 500.
    2. Word length will be at least 1 and does not exceed 500.
    3. Each word contains only lowercase English alphabet a-z.

    Example 1:

    Input:
    [
      "abcd",
      "bnrt",
      "crmy",
      "dtye"
    ]
    
    Output:
    true
    
    Explanation:
    The first row and first column both read "abcd".
    The second row and second column both read "bnrt".
    The third row and third column both read "crmy".
    The fourth row and fourth column both read "dtye".
    
    Therefore, it is a valid word square.
    

    Example 2:

    Input:
    [
      "abcd",
      "bnrt",
      "crm",
      "dt"
    ]
    
    Output:
    true
    
    Explanation:
    The first row and first column both read "abcd".
    The second row and second column both read "bnrt".
    The third row and third column both read "crm".
    The fourth row and fourth column both read "dt".
    
    Therefore, it is a valid word square.
    

    Example 3:

    Input:
    [
      "ball",
      "area",
      "read",
      "lady"
    ]
    
    Output:
    false
    
    Explanation:
    The third row reads "read" while the third column reads "lead".
    
    Therefore, it is NOT a valid word square.

    Runtime: 19ms
     1 class Solution {
     2 public:
     3     bool validWordSquare(vector<string>& words) {
     4         if (words.empty()) return true;
     5         
     6         int m = words.size(), n = words[0].size();
     7         if (m != n) return false;
     8         for (int i = 0; i < m; i++) {
     9             // get the i-th column string
    10             string columnString;
    11             for (int j = 0; j < m; j++) {
    12                 if (words[j].size() > i)
    13                     columnString += words[j][i];
    14             }
    15             if (columnString != words[i]) return false;
    16         }
    17         return true;
    18     }
    19 };
  • 相关阅读:
    SQL 游标示例
    在与SQL Server 建立 连接时出现与网络相关的或特定于实例的错误。未找到或无法访问服务器
    Jquery 设置焦点
    MVC Hidden用法
    Jquery根据name取得所有选中的Checkbox值
    MVC程序部署后页面指向login.aspx
    ref 和out 关键字
    认识关系型数据库
    抽象类和接口详解
    1.穷举法
  • 原文地址:https://www.cnblogs.com/amazingzoe/p/6025920.html
Copyright © 2020-2023  润新知