需要解释此代码如何处理 Arraylist 值 [英] Need explanations on how this code processes Arraylist values

问题描述

public static void mystery1(ArrayList<Integer> list) { 
for (int i = list.size() - 1; i > 0; i--) { 
    if (list.get(i) < list.get(i - 1)) { 
        int element = list.get(i); 
        list.remove(i); 
        list.add(0, element); 
    } 
} 
System.out.println(list);
}

我正在尝试使用给定的 ArrayList 值 集来解决此方法的输出问题.在查看它之后，我很确定这段代码会将一对的较小值重新定位在列表的前面.我使用值 [2, 6, 1, 8] 的 ArrayList 进行输入，结果是 [1, 2, 6, 8]，正如我所料.但是对于 [30, 20, 10, 60, 50, 40] 它产生了 [10, 30, 40, 20, 60, 50] 当我期望 [40,50,10,20,30,60].因此，任何人都可以向我解释这段代码实际上是如何处理 Arraylist 的吗?谢谢！

I'm trying to solve the for the output of this method with given sets of ArrayList values . After looking over it, I was pretty sure this code would relocate the smaller value of a pair in the front of the list. I used an ArrayList of values [2, 6, 1, 8] for input and the result was [1, 2, 6, 8], which was as I expected. But for [30, 20, 10, 60, 50, 40] it produced [10, 30, 40, 20, 60, 50] when I expected [40,50,10,20,30,60]. Therefore can anyone please explain to me how this code actually processes the Arraylist ? Thanks!

推荐答案

这是发生的事情，一步一步:

Here is what happens, step by step:

• 列表的初始状态:[30, 20, 10, 60, 50, 40]
• 是 40 <50?是的.所以去掉40，把它作为列表的第一个元素:[40, 30, 20, 10, 60, 50]
• 是 60 <10?不.不要碰任何东西:[40, 30, 20, 10, 60, 50]
• 是 10 <20?是的.所以去掉10，把它作为列表的第一个元素:[10, 40, 30, 20, 60, 50]
• 是 30 <40?是的.所以去掉30，把它作为列表的第一个元素:[30, 10, 40, 20, 60, 50]
• 是 10<30?是的.所以删除 10，并将其作为列表的第一个元素:[10, 30, 40, 20, 60, 50].

结果是:[10, 30, 40, 20, 60, 50]

The result is then: [10, 30, 40, 20, 60, 50]

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