在JavaScript中使用循环查找连续整数的总和 [英] Find the sum of consecutive whole numbers w/o using loop in JavaScript

问题描述

我正在寻找一种方法来进行如下计算：

I'm looking for a method to do calculations like:

function sumIntegerUpTo(number) {
  return 1+2+3+...+number;
}

如果你传递数字 as 5 函数应返回 1 + 2 + 3 + 4 + 5 的总和。我想知道是否可以不循环。

If you pass number as 5 function should return the sum of 1+2+3+4+5. I'm wondering if it's possible to do without loops.

推荐答案

function sumIntegerUpTo(number) {
    return (1 + number) * number / 2;
}

我可以想到两种简单的方法来记住这个公式：

I can think of two easy ways for me to remember this formula:

• 考虑从序列的两端添加数字：1和n，2和n-1,3和n-2，这些小额中的每一个最终都等于n + 1。两端将在序列的中间（平均）结束，因此总共应该有n / 2个。所以sum =（n + 1）*（n / 2）。

• Think about adding numbers from both ends of the sequence: 1 and n, 2 and n-1, 3 and n-2, etc. Each of these little sums ends up being equal to n+1. Both ends will end at the middle (average) of the sequence, so there should be n/2 of them in total. So sum = (n+1) * (n/2).

平均值之前有多少个数（即（1 + n）/ 2）因为有后，并且添加一对与该平均值等距的数字总是导致平均值的两倍，并且有n / 2对，所以sum =（n + 1）/ 2 * 2 * n / 2 =（n + 1）/ 2 * n。

There are as many number before the average (which is (1+n)/2) as there are after, and adding a pair of numbers that are equidistant to this average always results in twice the average, and there are n/2 pairs, so sum = (n+1)/2 * 2 * n/2 = (n+1)/2*n.

您可以相当轻松地将上述推理扩展到另一个起点数字，给你：总和（从a到b的数字，包括在内）=（a + b）/ 2 *（b-a + 1）。

You can fairly easily extend the above reasoning to a different starting number, giving you: sum(numbers from a to b, inclusive) = (a+b)/2*(b-a+1).

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