Codewars 中的“Yes No Yes No”任务

Question

https://www.codewars.com/kata/573c84bf0addf9568d001299/train/python

任务：

“编写一个接收数字或字符串数组的代码，一个接一个地通过它，同时取出一个值，留下一个值，取，离开，然后再次回到开头，直到所有值都出来。

这就像一群人决定每两个人都会离开它，直到最后一个人在那里。因此，如果数组的最后一个元素被取出，则仍然存在的第一个元素将保留。

该代码返回一个新的重新排列的数组，其中包含按顺序排列的已取值。始终采用初始数组的第一个值。

例子：

var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

// returns [1, 3, 5, 7, 9, 2, 6, 10, 8, 4]

var arr = ['this', 'code', 'is', 'right', 'the']

// returns ['this', 'is', 'the', 'right', 'code']

我的代码是：

def yes_no(arr):

    arr1 = []

    if len(arr) == 0:

        return arr1

    for i in range(len(arr)):

        if i % 2 == 0:

            arr1.append(arr[i])

    for j in arr1:

        arr.remove(j)

    yes_no(arr)

Answer 1

你不能这样做，因为可能有重复的数字，比如一个示例案例。.arr.remove(j)[1,2,3,4,5,20,6,7,8,20]我解决了这个问题，但是既然你提到了标签，答案可以是.javascriptalgorithmlanguage-agnostic方法：我们可以创建给定数字的循环双链表。因此，对于 ，列表将如下所示：[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; _____&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;|&nbsp; &nbsp; &nbsp;|&nbsp;&nbsp; &nbsp; &nbsp; both next and prev links(double arrow notation)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; v&nbsp; &nbsp; &nbsp;|&nbsp; -1 <--> 2 <--> 3 <--> 4 <--> 5 <--> 6 <--> 7 <--> 8 <--> 9 <--> 10 --&nbsp; | prev link&nbsp;| ^&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; |&nbsp; |&nbsp;| |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _next link_ _ |&nbsp; |&nbsp;|_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|对于每个步骤，我们将迭代中节点的当前值添加到结果中。在转到下一个节点之前（因为我们在中间跳过），我们将执行以下2个步骤：&nbsp; temp.prev.next = temp.next;&nbsp; temp.next.prev = temp.prev;这意味着我们将前一个节点的值分配给当前节点的下一个节点，将下一个节点的前一个值分配给当前节点的前一个值。next在迭代的第一步之后，我们的新（如修改后的）循环 DLL 将如下所示：&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;______&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; |&nbsp; &nbsp; &nbsp; |&nbsp;&nbsp; &nbsp; both next and prev link&nbsp; &nbsp; &nbsp;v&nbsp; &nbsp; &nbsp; |&nbsp; -2 <--> 4 <--> 6 <--> 8 <--> 10 --&nbsp; &nbsp;|&nbsp;| ^&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; |&nbsp; |prev link&nbsp;| |_ _ _ _ _ _ _ _ _next link_ _ _ |&nbsp; |&nbsp;|_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|同样，您可以生成每个步骤的列表外观。片段：function yesNo(arr) {&nbsp; var result = [];&nbsp; var head = null,&nbsp; &nbsp; temp = null;&nbsp; for (var i = 0; i < arr.length; ++i) {&nbsp; &nbsp; if (head == null) {&nbsp; &nbsp; &nbsp; head = createNode(arr[i]);&nbsp; &nbsp; &nbsp; temp = head;&nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; var temp2 = createNode(arr[i]);&nbsp; &nbsp; &nbsp; temp.next = temp2;&nbsp; &nbsp; &nbsp; temp2.prev = temp;&nbsp; &nbsp; &nbsp; temp = temp2;&nbsp; &nbsp; }&nbsp; }&nbsp; temp.next = head;&nbsp; head.prev = temp;&nbsp; temp = head;&nbsp; while (temp.next != temp) {&nbsp; &nbsp; result.push(temp.value);&nbsp; &nbsp; temp.prev.next = temp.next;&nbsp; &nbsp; temp.next.prev = temp.prev;&nbsp; &nbsp; temp = temp.next.next; // skip next and go to next to next&nbsp; }&nbsp; result.push(temp.value);&nbsp; return result;}function createNode(val) {&nbsp; return {&nbsp; &nbsp; value: val,&nbsp; &nbsp; prev: null,&nbsp; &nbsp; next: null&nbsp; };}console.log(yesNo([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])); console.log(yesNo(['this', 'code', 'is', 'right', 'the']));时间复杂度：O（n）空间复杂度：O（n）其中 是数组中的元素数。n

Answer 2

您可以使用 deque（从集合中）执行此操作，方法是在队列中弹出和旋转条目：from collections import dequearr&nbsp; &nbsp; = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]result = [ (q.popleft(),q.rotate(-1))[0] for q in [deque(arr)] for _ in arr]输出：[1, 3, 5, 7, 9, 2, 6, 10, 8, 4]您还可以创建一个函数，该函数将按正确的顺序计算索引，并在以下索引处返回元素：def skipOrder(arr,skipBy=1):&nbsp; &nbsp; N = len(arr)&nbsp; &nbsp; b = 2**N-1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;# bits of unskipped posiitons&nbsp; &nbsp; pos = skip = 0&nbsp; &nbsp; &nbsp;# current position and skip cycle&nbsp; &nbsp; while b:&nbsp; &nbsp; &nbsp; &nbsp; if b&1:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; # remaining position&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if not skip:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;# yield and clear if not skipped&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; b ^= 1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; yield arr[pos]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; skip = (skip+1)%(skipBy+1) # cycle skipping&nbsp; &nbsp; &nbsp; &nbsp; b&nbsp; &nbsp;= ((b&1)<<(N-1)) | (b>>1)&nbsp; # rotate positions bits&nbsp; &nbsp; &nbsp; &nbsp; pos = (pos+1)%N&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; # and indexresult = list(skipOrder(arr)) # [1, 3, 5, 7, 9, 2, 6, 10, 8, 4]它使用与队列类似的原理（屈服，跳过，旋转），但在数字的位上执行此操作，而不是移动数据结构中的实际元素。