Question

当每个要素只是一个扼杀时,我如何能够从关键因素中获得价值?

我的守则是:

public List<int[]> FindAllLetters(List<string> lettersOnBoard, Dictionary<string, int> frequency)
{
        List<int[]> lettersExistLeft = new List<int[]>();

        foreach (string letter in frequency.Keys)
        {
            int exist = 0;
            int dictionaryFrquencyLetter = 0;

            exist = lettersOnBoard.Count(n => n == letter);
            dictionaryFrquencyLetter = letter.GiveValueOfKey() //< how get value of key(letter) ?

        }
        throw new NotImplementedException();
}

是否可能这样做?

Answer 1

你们是否试图这样做?

dictionaryFrquencyLetter = frequency[letter];

Answer 2

一位具有一定关键意义的独裁者也给予你同样的重视:

dictionaryFrquencyLetter = frequency[letter];

然而(假设你所张贴的相当毫无意义的法典在某个时候会有所助益),如果你通过收集字典钥匙来收集,为什么不仅通过这些项目,例如项目,加以篡改。

foreach (var item in frequency)
{
    var key = item.Key
    var value = item.Value
}

Answer 3

It sounds like you are looking for this:

dictionaryFrquencyLetter = frequency[letter];

友情链接