随机排列 n 个元素的数组 a C++

Question

我对随机数组有疑问问题是，根据随机种子，结果又是相同的，而不是 2-exchange，两个数组是相同的！我想要 2 个结果数组随机交换

代码为2-exchange

#include <stdio.h>
#include <cstdlib>
#include <ctime>

void swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}

void printArray(int arr[], int n)
{
    for(int i = 0; i < n; i++)
        printf("%d ", arr[i]);
    printf("\n");
}

// A function to generate a random exchange
int* randomized(int arr[], int n)
{
    // Use a different seed value so that we don't get same
    // result each time we run this program

    srand(time(NULL));
    // Start from the last element and swap one by one. We don't
    // need to run for the first element that's why i > 0
    for(int i = n - 1; i > 0; i--)
    {
        // Pick a random index from 1 to i-1
        int j = rand() % (i - 1 + 1) + 1;
        //int j = rand() % (i+1);

        // Swap arr[i] with the element at random index
        swap(&arr[i], &arr[j]);
    }

    return arr;
}

// Driver program to test above function.
int main()
{
    int *x1, *x2;
    int arr[] = {6, 1, 2, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    x1 = randomized(arr, n);
    x2 = randomized(x1, n);
    printArray(x1, n);
    printArray(x2, n);
    getchar();
}

Answer 1

显然，randomized 在这两种情况下都 return 将指针指向传入的同一数组 (x1 = x2)。 "randomized" 听起来像是一个函数的名称，它应该 return 随机复制数据，而不是修改原始数据数据.

确定您想要的版本，并根据该版本修复功能或测试程序。

随机排列 n 个元素的数组 a C++

To shuffle an array a of n elements C++

c++

arrays

algorithm

pointers

shuffle