Question

因此，我们的目标是将一个单词整理成字母，并根据每个字母值给它一个分数。我有这个工作，并把它作为一个字典使用得分作为关键字和单词作为值。我需要找到仅使用所提供的字母的最高得分词（每个只使用一次），如果有多个得分相同，则需要打印添加到词典中的第一个。代码如下;

    Dictionary<string, int> wordHolder = new Dictionary<string, int>();



    int N = int.Parse(Console.ReadLine());
    for (int i = 0; i < N; i++)
    {
        string W = Console.ReadLine();


        char[] splitWord = W.ToCharArray();
        Console.Error.WriteLine(W);


        int points = splitWord.Sum(c => letterPoints.First(kvp => kvp.Value.Contains(c)).Key);
        wordHolder.Add(W, points);
        Console.Error.WriteLine(points);

    }
    string LETTERS = Console.ReadLine();
    Console.Error.WriteLine(LETTERS);

字母是单个字符串中提供的字符

Answer 1

您可以通过降序排序并获得第一个元素：

wordHolder.OrderByDescending(x => x.Value).First();

或按升序排序并获取最后一个元素：

wordHolder.OrderBy(x => x.Value).Last();

Answer 2

using System;
using System.Collections.Generic;
using System.Linq;

namespace ConsoleApplication9
{
  internal static class Program
  {
    private static void Main()
    {
      //  --- building up the data so the next (repeated) steps are efficient: ---

      // your words and points
      var wordHolder = new Dictionary<string, int>();

      // INSERT DATA INTO wordHolder HERE

      // create a dictionary where you can access all words and their points 
      // efficiently by the letters you have available:
      var efficientAccessByLetters = wordHolder.Select(entry =>
                                                new
                                                {
                                                  Index = new string(entry.Key.OrderBy(c => c).ToArray()),
                                                  Word = entry.Key,
                                                  Points = entry.Value
                                                })
                                                .GroupBy(x => x.Index)
                                                .ToDictionary(x => x.Key,
                                                              x => x.OrderByDescending(p => p.Points).ToList());

      //  --- Repeat this steps as many times as you like: ---

      while (true)
      {
        Console.WriteLine("Enter letters available:");

        // get letters that you want to build he best word from:
        var availableLetters = Console.ReadLine();

        // normalize the letters for fast index search by sorting them
        var normalized = new string(availableLetters.OrderBy(c => c).ToArray());

        if (!efficientAccessByLetters.ContainsKey(normalized))
        {
          Console.WriteLine("Sorry, no matching words found.");
        }
        else
        {
          // find all words and their respective points by index access
          var wordsThatMatchLetters = efficientAccessByLetters[normalized];

          // as words are sorted by points, just get the first
          var best = wordsThatMatchLetters.First();

          // output
          Console.WriteLine("Best Match: {0} for {1} points", best.Word, best.Points);
        }
      }
    }
  }
}

Answer 3

无需字典即可做到的最佳方法。

def save(filename, budgets):
   """Saves the changes made into the accounts"""

   with open(filename, 'w') as f_obj: 
       budgets2 = [budget.__dict__ for budget in budgets]
       json.dump(budgets2, f_obj)

使用键从字典中选取最高值

3 个答案: