重新排列数组以匹配位置数组 [英] Re-arrange the array to match the location array

问题描述

给定一个整数数组和一个具有这些整数位置的数组。如何重新排列数字以匹配将数据放入新订单中



ie：



阵列1,2,4



位置阵列1,2,0



数据第三个位置应该移到第一个位置（4）

第一个位置的数据应该移到第二个位置（1）

第二个位置的数据应该移动到第三个地方（2）



我尝试过：

Given an array of integers and an array that has the location of those integers. How do I re-arrange the numbers to match put the data in the new order

i.e:

an array 1,2,4

location array 1,2,0

data in the third spot should be moved to the first spot (4)
data in the first spot should be moved to the second (1)
data in the second spot should be moved to the third spot (2)

What I have tried:

#include <iostream>
using namespace std;



// assign the array to have a maximum of 3 integer values 
const int MAX = 3;
 
int main ()
{
   
   // the elements in the array (the three elements allowed) 
   int  var[MAX] = {10, 100, 200};
   
   // declares ptr as an array of MAX integer pointers
   // each element in ptr, now holds a pointer to an int value
   int *ptr[MAX];
 
   // for loop increments until it hits the amount of elements in the array 
   // in this case it will loop three times starting at the 0 index and ending at 2 index (3 elements)
   // three integers which will be stored in an array of pointers 
   for (int i = 0; i < MAX; i++)
   {
      // assign the address of integer. 
      ptr[i] = &var[i]; 
   }
   
   // prints out the elements location, i, as well as its value *ptr
   for (int i = 0; i < MAX; i++)
   {
      cout << "Value of var[" << i << "] = ";
      cout << *ptr[i] << endl;
   }
   return 0;
}

推荐答案

试试这个：

Try this:

#include <iostream>
using namespace std;
 
// assign the array to have a maximum of 3 integer values 
const int MAX = 3;
 
int main ()
{
   // the elements in the array (the three elements allowed) 
   int  var[MAX] = {10, 100, 200};
   // position array
   int  pos[MAX] = {1, 2, 0};
   // destination array
   int  dest[MAX];
   for (int i = 0; i < MAX; i++)
   {
      dest[pos[i]] = var[i]; 
   }
   
   // prints out the elements location, i, as well as its value *ptr
   for (int i = 0; i < MAX; i++)
   {
      cout << "Value of dest[" << i << "] = ";
      cout << dest[i] << endl;
   }
   return 0;
}</iostream>

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