找到两个数组的交集

My aim is to find out values of intersection of arrays a and b and store them into a new array c so the printout will be : 3,10,4,8. How do I assign given values to a 3rd array c ?

我的目标是找出数组a和b的交集值并将它们存储到一个新的数组c中，因此打印输出将为：3,10,4,8。如何将给定值分配给第3个数组c？

 public static void main(String[] args) {
        int a[] = {3, 10, 4, 2, 8};
        int[] b = {10, 4, 12, 3, 23, 1, 8};
        int[] c;
        int i=0;
         for(int f=0;f<a.length;f++){
              for(int k=0;k<b.length;k++){
                    if(a[f]==b[k]){
 //here should be a line that stores equal values of 2 arrays(a,b) into array c
            }
          }
        }
            for (int x=0; x<c.length; x++){
             System.out.println(c[i]);
            }
       }
  }

5 个解决方案

#1

This should be an easy way to do.

这应该是一个简单的方法。

int a[] = {3, 10, 4, 2, 8};
int[] b = {10, 4, 12, 3, 23, 1, 8};
List<Integer> aList =  Arrays.asList(a);
List<Integer> bList =  Arrays.asList(b);
aList.retainAll(bList);
System.out.println(" a intersection b "+aList);
int[] c = aList.toArray(new int[0]);

#2

public static void main(String[] args) {
        int a[] = {3, 10, 4, 2, 8};
        int[] b = {10, 4, 12, 3, 23, 1, 8};
        int[] c = new int[(int)Math.min(a.length, b.length)];
        int i=0;
         for(int f=0;f<a.length;f++){
              for(int k=0;k<b.length;k++){
                    if(a[f]==b[k]){
                    c[i] = a[f];
                    i++;
            }
          }
        }
        for (int x=0; x<i; x++){
           System.out.println(c[x]);
        }
       }
  }

Hope it helps. Or if you have time complexity issue then try Java Set.

希望能帮助到你。或者，如果您有时间复杂性问题，请尝试Java Set。

#3

First you need to allocate space for your array:

首先，您需要为数组分配空间：

int[] c = new int[SOME_SIZE];

The hard part is figuring out how much SOME_SIZE should be. Since you are calculating an intersection, the most it can be is the size of the smallest of a and b.

困难的部分是弄清楚SOME_SIZE应该是多少。由于您正在计算交叉点，因此最大可能是a和b中最小的交叉点的大小。

Finally, to assign an element in the array, you just do

最后，要在数组中分配一个元素，你就可以了

c[idx] = a[f]

Now you need to keep track of where idx goes. I suggest starting with idx = 0 and incrementing it each time you find a new element to add to c.

现在你需要跟踪idx的去向。我建议从idx = 0开始并在每次找到要添加到c的新元素时递增它。

#4

if permitted use ArrayList for c, its growable array

如果允许，则使用ArrayList表示c，其可扩展数组

ArrayList c = new ArrayList();
.
.
.
.
.
c.add(a[f]);

also if permitted to sort arrays, i recommend you to sort smaller array and then iterate over larger array and binary search in smaller array.

如果允许对数组进行排序，我建议您对较小的数组进行排序，然后在较小的数组中迭代较大的数组和二进制搜索。

#5

You can take a help of temporary variable (but this is basically reinventing the wheel, if you are not required to do so) -

你可以采取临时变量的帮助（但如果你不需要这样做，这基本上是重新发明*） -

int[] c = new int[0];
//...
    if(a[f] == b[k]) { 
        int[] temp = c;
        c = new int[c.length + 1];
        for(int i=0; i<temp.length; i++) {
            c[i] = temp[i];
        }
        c[c.length - 1] = a[f];
    }
//...

#1