更改您的第二个为循环：

 的for（int i = N-1; I＆GT; = 0; --i）{
    //生成一个随机数[0，N-1]
    INT J =兰特（）％（I + 1）;    //交换最后一个元素与元素随机指数
    INT温度= R [I]
     -  [R [i] = R [J]。
     -  [R [J] =温度;
}
 

这是一个费雪耶茨洗牌算法。我一直在使用听说兰特（）％N 不均匀分布，你已经被警告。

如果你想每次生成唯一的排列，可以存储生成的排列，可能在词典或的Hashmap ，然后查找每次你回来。我不认为 C 有一个内置的之一，但应该有可用的库。

I've written this function in C and I want it to create a random permutation or a list of numbers from 1 to n. I'm having trouble getting it to have no repeating numbers. So if you have n = 4, i would like it to return a random array containing 1-4 each only once, for example: {1,3,4,2}

int* random(int n) 
{
    int* r = malloc(n * sizeof(int));
    // initial range of numbers
    for(int i=0;i<n;++i){
        r[i]=i+1;
    }
    // shuffle
    for (int i = 1; i <= n; ++i){
        int j = rand() % i;
        r[i] = r[j];
        r[j] = i;
  }
  return r;
}

Change your second for loop to:

for (int i = n-1; i >= 0; --i){
    //generate a random number [0, n-1]
    int j = rand() % (i+1);

    //swap the last element with element at random index
    int temp = r[i];
    r[i] = r[j];
    r[j] = temp;
}

This is a Fisher-Yates shuffling algorithm. I have heard using rand() % n doesn't distribute uniformly, you've been warned.

And if you want to generate unique permutations everytime, you can store the generated permutations, maybe in a Dictionary or Hashmap, and then lookup everytime you return. I don't think C has a built in one but there should be libraries available.