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重新排列一個數組,使得每個奇數索引的元素都大於其前一個元素

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2023-09-01 09:45:071171瀏覽

重新排列一個數組,使得每個奇數索引的元素都大於其前一個元素

我們有一個正整數型別的數組,假設為arr[],大小任意。任務是重新排列數組,使得所有奇數索引位置的元素的值大於偶數索引位置的元素,並列印結果。

讓我們來看看各種輸入輸出狀況:

輸入 − int arr[] = {2, 1, 5, 4, 3, 7, 8}

輸出 − 排列前的陣列:2 1 5 4 3 7 8 將陣列重新排列,使得每個奇數索引的元素都大於其前一個元素:1 4 2 5 3 8 7

解釋  - 我們給定一個大小為7的整數數組。現在,如果偶數索引的元素較大,我們將交換偶數索引處的元素和奇數索引處的元素

Arr[0] > arr[1] = call swap = {1, 2, 5, 4, 3, 7, 8}
Arr[2] > arr[3] = call swap = {1, 2, 4, 5, 3, 7, 8}
Arr[6] > arr[5] = call swap = {1, 2, 4, 5, 3, 8, 7}
Arr[2] > arr[1] = call swap = {1, 4, 2, 5, 3, 8, 7}

輸入− int arr[] = {3, 2, 6, 9}

輸出− 排列前的陣列: 3 2 6 9 Rearrangement of an array such that every odd indexed element is greater than it previous is: 2 3 6 9

Explanation − we are given an integer array of size#Explanation

− we are given an integer array of size 4. Nowwe the elements at even index with the elements at odd index if even indexed elements are greater i.e. Arr[0] > arr[1] = call swap = {2, 3, 6, 9}. No need to further call the swap = {2, 3, 6, 9}. No need to further call the swap method the swap method as all the elements at positions satisfies the conditions

Approach used in the below program is as follows
  • Input an array of integer type elements and calculate the
  • #Input an array of integer type elements and calculate the ##size of an array .
  • Print the array before arrangement and call the function Rearrangement(arr, size)
    • Inside the function Rearrangement(arr, size)
    • Create a variable of integer type let's say, ptr and set it with size-1.
    • Start loop FOR, from i to 0 till i less than ptr and i = i 1. Inside the loop, check if arr[i] is greater than arr[i 1] then call swap(arr[i], arr[i 1]).
    • Check IF size & 1 then start loop FOR from i to ptr till i greater than 0 and i = i - 2. Inside the loop, check IF arr[i] greater than arr[i - 1] then call swap(arr[i], arr[i-1])
  • Print the array after the rearrangement of values of an array.

#Example

#include <iostream>
using namespace std;
void Rearrangement(int arr[], int size){
   int ptr = size - 1;
   for(int i = 0; i < ptr; i = i+2){
      if(arr[i] > arr[i+1]){
         swap(arr[i], arr[i+1]);
      }
   }
   if(size & 1){
      for(int i = ptr; i > 0; i = i-2){
         if(arr[i] > arr[i-1]){
            swap(arr[i], arr[i-1]);
         }
      }
   }
}
int main(){
   //input an array
   int arr[] = {2, 1, 5, 4, 3, 7, 8};
   int size = sizeof(arr) / sizeof(arr[0]);
   //print the original Array
   cout<<"Array before Arrangement: ";
   for (int i = 0; i < size; i++){
      cout << arr[i] << " ";
   }
   //calling the function to rearrange the array
   Rearrangement(arr, size);
   //print the array after rearranging the values
   cout<<"\nRearrangement of an array such that every odd indexed element is greater than it previous is: ";
   for(int i = 0; i < size; i++){
      cout<< arr[i] << " ";
   }
   return 0;
}

輸出

如果我們執行上述程式碼,將會產生以下輸出###
Array before Arrangement: 2 1 5 4 3 7 8
Rearrangement of an array such that every odd indexed element is greater than it previous is: 1 4 2 5 3 8 7
###

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