我怎样才能整理出这个容器 [英] How can I sort out this container

问题描述

我正在寻找一个最年轻的玩家，我需要对这个容器进行排序并选择第一个玩家，但是我不知道该怎么做，它说左手边必须是变量属性或索引器。



我尝试了什么：



我试过泡泡排序方法，但它不适合我或者我只是做错了，不知道。

I'm looking for a youngest player, i need to sort this container and take the the first player, however I dont know how to do it, it says "the left hand side must be a variable property or indexer".

What I have tried:

I've tried "bubble Sort" method, yet it did not work for me or maybe im just doing it wrong, no idea.

<pre>        public Player FindYoungestPlayer(PlayersContainer entireSportsTeam)
        {
            Player youngestPlayer = entireSportsTeam.GetPlayer(0);
            for (int i = 0; i < entireSportsTeam.Count; i++)
            {
                for (int j = 0; j < entireSportsTeam.Count-1; j++)
                {
                    if (entireSportsTeam.GetPlayer(i).Birthday > youngestPlayer.Birthday)
                    {
                        youngestPlayer = entireSportsTeam.GetPlayer(i);
                        entireSportsTeam.GetPlayer(i) = entireSportsTeam.GetPlayer(j);
                        entireSportsTeam.GetPlayer(j) = youngestPlayer;
                    }
                }
            }
            return youngestPlayer;
        }

推荐答案

问题是你没有正确交换，因为你的比较不是你想象的那样是的，即使它匹配，j与你想要交换的东西无关。



如果你想找到最年轻的，那只是一次通过 - 你根本不需要排序或使用嵌套循环：

The problem is that you aren't swapping properly, because your comparison isn't what you think it is, and even when it matches, j is irrelevant to what you want to swap.

If you want to find the youngest, that that's a single pass though - you don't need to sort, or use a nested loop at all:

Player youngestPlayer = entireSportsTeam.GetPlayer(0);
for (int i = 0; i < entireSportsTeam.Count; i++)
{
    if (entireSportsTeam.GetPlayer(i).Birthday > youngestPlayer.Birthday)
    {
        youngestPlayer = entireSportsTeam.GetPlayer(i);
    }
}

并且您不应该在名为FindXXX的方法中对任何内容进行排序，因为该名称意味着根本不会对数据进行任何更改。

And you shouldn't sort anything in a method called "FindXXX" because the name implies no changes to data at all.

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