由网友(沉醉美色。)分享简介：我试图解决以下问题。有两个数组大小n和大小为n + 1的B。 A和B的所有元素相同。 B有一个额外的元素。找到的元素。I am trying to solve the following problem. There are two arrays A of size n and B of size n+1. A a...

我试图解决以下问题。有两个数组大小n和大小为n + 1的B。 A和B的所有元素相同。 B有一个额外的元素。找到的元素。

I am trying to solve the following problem. There are two arrays A of size n and B of size n+1. A and B have all elements same. B has one extra element. Find the element.

我的逻辑是将数组转换为列表并检查B中每个元素是present于A。

My logic is to convert the array to list and check if each element in B is present in A.

但是，当我使用的基本数组我的逻辑是行不通的。如果我使用

But when I am using the primitive arrays my logic is not working. If I am using

Integer [] a ={1,4,2,3,6,5};
Integer [] b = {2,4,1,3,5,6,7};

我的code是工作的罚款。

My code is working fine.

public static void main(String [] args)
{
    int [] a ={1,4,2,3,6,5};
    int [] b = {2,4,1,3,5,6,7};     
    List<Integer> l1 = new ArrayList(Arrays.asList(a));
    List<Integer> l2 = new ArrayList(Arrays.asList(b));
    for(Integer i :l2)
    {
        if(!l1.contains(i))
        {
            System.out.println(i);
        }           
    }
}

也是我的逻辑是O（n + 1）。有没有更好的算法中。

And also My logic is O(n+1). Is there any better algo.

感谢

推荐答案

这是不工作的基本数组的原因，就是 Arrays.asList 定当 INT [] 返回列表＆LT;整数[]> 而不是一个名单，其中，整型>

The reason it is not working for primitive arrays, is that Arrays.asList when given an int[ ] returns a List<Integer[ ]> rather than a List<Integer>.

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