Informatica 编码轮问题

给定一个包含 n 个元素和一个整数 k 的数组。将数组划分为子数组，每个子数组包含k个元素。例如：
输入：arr[]={1, 32, 5, 6, 9, 3} 和 k=2

子数组将包含元素：
{132}、{56}、{93}。

现在将这些子数组排序为 {56}、{93}、{132}。将这些子数组合并在一起，并按排序顺序将它们显示为原始数组的元素
最终输出：arr[]={5, 6, 9, 3, 1, 32}

例子：

Input : 
arr[]={1, 32, 5, 6, 9, 3} 
k=2 
Output : 
arr[]={5, 6, 9, 3, 1, 32}

下面是上述思想的实现：

CPP

// C++ Program implementation of the above idea
 
#include 
using namespace std;
void subarray_k(vector v, int n, int k)
{
    int sum = 0;
    int index;
    string str = "";
    vector > arr;
    for (int i = 0; i < n - 1; i++)
    {
        index = i;
        str = to_string(v[i]);
        i++;
        for (int j = 0; j < k - 1; j++)
        {
            // Conversion of array elements(integers) to
            // string for easy concatenation
            string temp = to_string(v[i]);
            str += temp;
            if (j != k - 2)
                i++;
        }
        stringstream check(str);
        int s = 0;
 
        // Conversion of concatenated string to integer
        check >> s;
        arr.push_back(make_pair(s, index));
        index++;
    }
    int u = 0;
   
    // Sorting of array
    sort(arr.begin(), arr.end());
    vector v2;
    for (int i = 0; i < (n / k); i++) {
        u = 0;
        int j = arr[i].second;
        while (u < k) {
            int a = v[j];
            v2.push_back(a);
            u++;
            j++;
        }
    }
   
    // print
    for (int i = 0; i < n; i++)
    {
        cout << v2[i] << " ";
    }
}
 
// Driver code
int main()
{
    int n = 6;
    vector v = { 1, 32, 5, 6, 9, 3 };
    int k = 2;
 
    // Function call
    subarray_k(v, n, k);
    return 0;
    // This code is contributed by Siddhant Thapliyal
}

输出

5 6 9 3 1 32