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Ростислав Юшкевич
Asked:2020-05-06 07:44:30 +0000 UTC2020-05-06 07:44:30 +0000 UTC

排序前打印数组元素的索引

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按降序对整数数组 X=(x1,x2,..,xn) 进行排序。按照元素最初的顺序打印原始数组元素的索引。打印数组中最大和最小元素的索引。这是代码：由于某种原因，如果您在排序之前查找数组的最大和最小元素，那么在它之后，数组将被第一个元素填充。以及如何以原始顺序显示索引？先感谢您

   #include <iostream>
#include <stdlib.h>
#include <stdio.h>

void shellsort(int* a, int leng);

int main()
{
    int n, i, j = 0;
    printf("Enter size:\n");
    scanf_s("%d", &n);
    printf("Enter elements: \n");
    int* mas = new int[n];
    for (i = 0; i < n; i++)
    {
        scanf_s("%d", &mas[i]);
    }
    int max = mas[0];
    int d = 0;
    for (i = 0; i < n; i++)
    {
        if (max < mas[i])
        {
            mas[i] = max;
            d = i;
        }
    }
    printf("Max index is: %d\n", d);
    int min = mas[0];
    int e = 0;
    for (i = 0; i < n; i++)
    {
        if (min > mas[i])
        {
            mas[i] = min;
            e = i;
        }
    }
    printf("Min index is: %d\n", e);
    shellsort(mas, n);
    printf("\n");

    delete[]mas;
}

void shellsort(int* a, int leng)
{
    int k = 0, i, j, p, temp, step;
    int* gap = new int[leng];
    gap[0] = leng / 2;
    while (gap[k] > 1)
    {
        k++;
        gap[k] = gap[k - 1] / 2;
    }
    for (i = 0; i <= k; i++)
    {
        step = gap[i];
        for (j = step; j < leng; j++)
        {
            temp = a[j];
            p = j - step;
            while (p >= 0 && temp > a[p])
            {
                a[p + step] = a[p];
                p = p - step;
            }
            a[p + step] = temp;
        }
    }
    printf("\n");
    for (i = 0; i < leng; i++)
    {
        printf("%d", a[i]);
    }
}
c++

1 个回答

  1. Best Answer

    我希望我正确理解了这种情况。这个想法是将原始索引保持在与元素相同的结构中。然后，在排序时，索引将按正确的顺序排列。

    #include <iostream>

void shellsort(std::pair<int, int>* a, int leng);

int main()
{
    int n, i, j = 0;
    std::cout << "Enter size:" << std::endl;
    std::cin >> n;
    std::cout << "Enter elements:" << std::endl;
    std::pair<int, int>* mas = new std::pair<int, int>[n];
    for (i = 0; i < n; i++)
    {
        std::cin >> mas[i].first;
        mas[i].second = i;
    }
    shellsort(mas, n);
    for (int i = 0; i < n; ++i) {
        std::cout << mas[i].first << " ";
    }
    std::cout << std::endl;
    for (int i = 0; i < n; ++i) {
        std::cout << mas[i].second << " ";
    }
    std::cout << std::endl;
    std::cout << "min ind: " << mas[n - 1].second << std::endl;
    std::cout << "max ind: " << mas[0].second;
    delete[] mas;
}

void shellsort(std::pair<int, int>* a, int leng)
{
    int k = 0, i, j, p, step;
    std::pair<int, int> temp;
    int* gap = new int[leng];
    gap[0] = leng / 2;
    while (gap[k] > 1)
    {
        k++;
        gap[k] = gap[k - 1] / 2;
    }
    for (i = 0; i <= k; i++)
    {

        step = gap[i];
        for (j = step; j < leng; j++)
        {
            temp = a[j];
            p = j - step;
            while (p >= 0 && temp.first > a[p].first)
            {
                a[p + step] = a[p];
                p = p - step;
            }
            a[p + step] = temp;
        }
    }
}

    https://ideone.com/WFJCrq

    • 1

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