Josephus Permutation

 https://www.codewars.com/kata/5550d638a99ddb113e0000a2

This problem takes its name by arguably the most important event in the life of the ancient historian Josephus: according to his tale, he and his 40 soldiers were trapped in a cave by the Romans during a siege.

Refusing to surrender to the enemy, they instead opted for mass suicide, with a twist: they formed a circle and proceeded to kill one man every three, until one last man was left (and that it was supposed to kill himself to end the act).

Well, Josephus and another man were the last two and, as we now know every detail of the story, you may have correctly guessed that they didn't exactly follow through the original idea.

You are now to create a function that returns a Josephus permutation, taking as parameters the initial array/list of items to be permuted as if they were in a circle and counted out every k places until none remained.

Tips and notes: it helps to start counting from 1 up to n, instead of the usual range 0..n-1; k will always be >=1.

For example, with n=7 and k=3 josephus(7,3) should act this way.

[1,2,3,4,5,6,7] - initial sequence
[1,2,4,5,6,7] => 3 is counted out and goes into the result [3]
[1,2,4,5,7] => 6 is counted out and goes into the result [3,6]
[1,4,5,7] => 2 is counted out and goes into the result [3,6,2]
[1,4,5] => 7 is counted out and goes into the result [3,6,2,7]
[1,4] => 5 is counted out and goes into the result [3,6,2,7,5]
[4] => 1 is counted out and goes into the result [3,6,2,7,5,1]
[] => 4 is counted out and goes into the result [3,6,2,7,5,1,4]

So our final result is:

josephus([1,2,3,4,5,6,7],3)==[3,6,2,7,5,1,4]

 约瑟夫问题，n个人围成一个圈，开始从1报数，123，123，123的这样报数，报到3的人被杀死，问最后剩下哪个人，需要将杀死的人的顺序输出。

我的算法：

缺陷：出局的人虽然被标记了，但是每次还得参与一次循环，到后期循环都是白走几次；

def josephus(items,k):
    result = []
    index = 0
    for _ in range(len(items)):
        count = 0
        while count

 

大神的算法：

很巧妙的将出局人的索引算出来了

def josephus(items,k):
    result = []    
    count = 0
    while len(items)>0:
        # 出局的人的索引是 i+k-1，因为需要循环报数，所以需要和数组长度求余
        count = (count + k -1)%len(items)
        result.append(items.pop(count))
    return result