fisher-yates_使用Fisher-Yates随机播放算法以O（n）时间随机播放给定数组

fisher-yates

Example:

例：

Say the input array is 
[1, 2 3, 4, 5 6, 7]
After reshuffling it can be anything like
[4, 3, 7, 2, 1, 5, 1]

Our goal is that the reshuffling should be as random as possible.

我们的目标是，改组应尽可能地随机。

There is a standard algorithm named Fisher-Yates algorithm which shuffles the array in linear times.

有一个名为Fisher-Yates算法的标准算法，可以在线性时间内对数组进行随机排序。

The idea is to start from an end

这个想法是从头开始

Say we are string from the right end and the array size is n

假设我们是从右端开始的字符串，数组大小为n

Now, pick the end element which will be the nth element, and pick up another element randomly from the range [0, n-1]. Then swap the elements and shrink the right end. Thus after this step, a[n-1](the rightmost element) is fixed. Continue the same until we reach the left end.

现在，选择将成为第n个元素的end元素，并从[0，n-1]范围内随机选择另一个元素。然后交换元素并缩小右端。因此，在此步骤之后， a [n-1] (最右边的元素)被固定。继续同样操作，直到到达左端。

The detailed algorithm will be,

详细的算法将是

For i=n-1 to 1
Pick and element in the range [0,i-1] randomly
Swap the randomly picked element with a[i]
Decrement i
End for loop

Since, it's random we can't do dry run

因为这是随机的，所以我们不能空运行

But still to illustrate lets pick elements randomly as per thought

但仍需说明让我们根据想法随机选择元素

So initially,
Array is
[1, 2 3, 4, 5 6, 7]
So i=6
Say, we picked up 4^th (0-indexed) element randomly (in the range [0-5])
Swap them
So array is now
[1, 2, 3, 4, 7, 6, 5]
So 5 is now fixed and is guaranteed to stay 
there after the whole shuffling completes
On the next iteration
i will be 5 and we will continue similar way until I becomes 1

C++ implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
//reference to array is passed as we need 
//to update the array within the function
void reshuffle(vector<int>& arr, int n)
{
srand(time(0));
for (int i = n - 1; i >= 1; i--) {
//j will be a random no with in range 0 to i-1
int j = rand() % i;
//swap ith index with jth
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
int main()
{
cout << "input array size\n";
int n;
cin >> n;
cout << "Input array elements \n";
vector<int> arr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
reshuffle(arr, n);
cout << "After reshuffling, printing the array\n";
for (auto it : arr)
cout << it << " ";
cout << endl;
return 0;
}

Output:

输出：

input array size
6
Input array elements
12 34 32 56 48 11
After reshuffling, printing the array
56 48 12 11 32 34

Real-life applications:

实际应用：

Creating an application which will suggest movie/songs from a list randomly. But each time it would suggest a unique movie only.

创建一个可以从列表中随机建议电影/歌曲的应用程序。但是每次都会只推荐一部独特的电影。

In this case, what we will do is we will show the i^th indexed movie after each iteration. As the i^th indexed one is never going to come again in the reshuffling as that's being fixed every time. That means we will recommend ith song/movie after the swap is done at each stage.

在这种情况下，我们要做的是在每次迭代后显示第i ^个索引的电影。由于第i ^个被索引的索引永远不会在改组中再次出现，因为每次都固定。这意味着在每个阶段完成交换后，我们将推荐第i首歌曲/电影。

Go through the below link to understand the detailed problem and solution.

浏览以下链接以了解详细的问题和解决方案。