Question

我一直试图完成这项工作，但我现在陷入了困境。任何人都可以推荐最简单的方法来完成它吗？谢谢

c[i] = a[i] + b[i];

// The snippet of code I have above works fine with operator: + - / * 
// a[i] = { 1, 8, 13, 15 }
// b[i] = { 3, 8, 13, 20 }
// then  c[i] = { 4, 16, 26, 35 }

// But how do i find the union and intersection of both arrays?
// I want the results to look like:
// union = {1,3,8,13,15,20}
// intersection = {8,13}

Answer 1

因为你要求最简单的方法。你可以把数组的元素放到2 std :: set

中

std::set<T> set1(begin(array1), end(array1));
std::set<T> set2(begin(array2), end(array2));

然后在2个新的std :: set中使用std :: set_union和std :: set_intersection。请查看在set union和set intersection这些链接中的完成方式因此，STL负责处理大多数逻辑和容器。

Answer 2

试试这个..

//Unioin of 2 array

#include<stdio.h>

int printUnion(int arr1[], int arr2[], int m, int n)
{
  int i = 0, j = 0;
  while(i < m && j < n)
  {
    if(arr1[i] < arr2[j])
      printf(" %d ", arr1[i++]);
    else if(arr2[j] < arr1[i])
      printf(" %d ", arr2[j++]);
    else
    {
      printf(" %d ", arr2[j++]);
      i++;
    }
  }


  while(i < m)
   printf(" %d ", arr1[i++]);
  while(j < n)
   printf(" %d ", arr2[j++]);
}


int main()
{
  int arr1[] = {1, 8, 13, 15};
  int arr2[] = {3, 8, 13, 20};
  int m = sizeof(arr1)/sizeof(arr1[0]);
  int n = sizeof(arr2)/sizeof(arr2[0]);
  printUnion(arr1, arr2, m, n);
  getchar();
  return 0;
}


//Intersection of two array

#include<stdio.h>

int printIntersection(int arr1[], int arr2[], int m, int n)
{
  int i = 0, j = 0;
  while(i < m && j < n)
  {
    if(arr1[i] < arr2[j])
      i++;
    else if(arr2[j] < arr1[i])
      j++;
    else /* if arr1[i] == arr2[j] */
    {
      printf(" %d ", arr2[j++]);
      i++;
    }
  }
}


int main()
{
  int arr1[] = {1, 8, 13, 15};
  int arr2[] = {3, 8, 13, 20};
  int m = sizeof(arr1)/sizeof(arr1[0]);
  int n = sizeof(arr2)/sizeof(arr2[0]);
  printIntersection(arr1, arr2, m, n);
  getchar();
  return 0;
}

最简单的方法是将两个双阵列组合成一个找到它们的联合和交叉点

2 个答案: