【5 kyu】Who Is Next？（ Double Cola ）

原題目

Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on.

For example, Penny drinks the third can of cola and the queue will look like this:

Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny

Write a program that will return the name of a man who will drink the n-th cola.

Note that in the very beginning the queue looks like that:

Sheldon, Leonard, Penny, Rajesh, Howard

題目： 有一個隊列Sheldon, Leonard, Penny, Rajesh, Howard，排到的人，再次排到隊尾，並將本身變成雙倍。若給了一個隊列names，排到的第r我的是誰？（題目比較繞，不知道該怎麼描述?）

// 初始隊列, 第一輪
["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]

// Sheldon排過隊以後，從新到隊尾，並將本身複製了一份
["Leonard", "Penny", "Rajesh", "Howard", "Sheldon", "Sheldon"]

// ...
// 第二輪，隊列以下
["Sheldon", "Sheldon", "Leonard", "Leonard", "Penny", "Penny", "Rajesh", "Rajesh", "Howard", "Howard"]

// 第三輪，隊列以下
["Sheldon", "Sheldon", "Sheldon", "Sheldon", "Leonard", "Leonard", "Leonard", "Leonard", "Penny", "Penny", "Penny", "Penny", "Rajesh", "Rajesh", "Rajesh", "Rajesh", "Howard", "Howard", "Howard", "Howard"]

My Solution

隊列每輪事後，隊列中的人數將變爲原來的二倍。若初始隊列的長度爲len，第r人位於第m輪

第一輪：n=1，sum = len
第二輪：n=2，sum = len + len * 2
第三輪：n=4，sum = len + len * 2 + len * 4
...
第m輪：n=2^m, sum = len + len * 2 + …… + len * n

則，第r我的在第m輪的排名爲：x = r + len * n - sum

則，第r我的的名字在初始隊列中排名爲t(數組下標爲t-1)，則：(t-1) * n < x <= t * n

因此，t = Math.ceil( x / n )

最後，代碼整理以下：

function whoIsNext(names, r){
  var len = names.length;
  var sum = names.length;
  var n = 1;
  while(sum < r) {
    n *= 2;
    sum += len * n;
  }
  return names[Math.ceil((r + len * n - sum) / n) - 1]
}

Best Practices / Clever

function whoIsNext(names, r) {
  var l = names.length;
  while (r >= l) { r -= l; l *= 2; }
  return names[Math.ceil(names.length * r / l)-1];
}

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對比

兩個方法一個用加法，一個用減法，得出最後一輪的人數。我的也以爲 減法 更 clever 一點 ?