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< N ++我){ - [R [i] = I + 1; } //洗牌 的for(int i = 1; I< = 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> = 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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