如何创建一个数组的随机排列？ [英] How to create a random permutation of an array?

问题描述

我用C编写的这个功能，我想它来创建一个随机排列或数字从1到n的列表。我无法得到它没有重复号码。所以，如果你有N = 4，我想它返回一个包含各1-4只进行一次随机序列，例如：{1,3,4,2}

 为int *随机（INT N）
{
    为int * R =的malloc（N * sizeof的（INT））;
    //号的初始范围
    的for（int i = 0; I＆LT; N ++我）{
         -  [R [i] = I + 1;
    }
    //洗牌
    的for（int i = 1; I＆LT; = N; ++ I）{
        INT J =兰特（）％I;
         -  [R [i] = R [J]。
         -  [R [J] =我;
  }
  返回ř;
}
 

解决方案

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;
}

解决方案

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.

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