如何创建一个数组的随机排列？

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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.