转换原始int数组列出 [英] Converting Primitive int array to list

问题描述

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>.

番石榴已经在Ints 的类。是有一个 asList 方法，将采取 INT [] ，然后返回一个列表＆LT;整数>

Guava has an answer to this in the Ints class. Is has an asList method that will take an int[ ] and return a List<Integer>

更新

int[] a = ...;
int[] b = ...;
List<Integer> aList = Ints.asList(a);
List<Integer> bList = Ints.asList(b);

以上将让您的code正常工作对于int []，因为它适用于整数[]。

The above will allow your code to work properly for int[ ] as it works for Integer[ ].

查看 INTS API

这篇关于转换原始int数组列出的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！