zoukankan      html  css  js  c++  java
  • 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:
    新版open live writer连接博客园
    github个人博客域名绑定(附带详细原理说明)
    开源性能监控分析工具glowroot
    (转)Python实例手册
    Jetty嵌入式Web容器攻略
    H2数据库攻略
    CAS ticket过期策略
    CAS自定义登录验证方法
    Sonatype Nexus高级配置
    配置sonar、jenkins进行持续审查
  • 原文地址:https://www.cnblogs.com/amazingzoe/p/6025920.html
Copyright © 2011-2022 走看看