In the second case, I found in the next smallest element and moved here, taking what was there and moving it on, in this case I would swap the 4 and the 8, and in next case I wouldn't have to do anything.

在第二次遍历中，我找到了,第二小的元素，把它移到这里,把这里原来的元素移到哪里,在这一次遍历中，我会把8和4交换,然后一次遍历，不会做任何事情。

麻省理工公开课 - 计算机科学及编程导论课程节选

Advance this pointer-- this finger, to the next element of the list which is 4, make the comparison.

前移这个指示器--也就是这根手指，让它指向列表中的,下一个元素4，再做比较。

哈佛公开课 - 计算机科学课程节选

s1 So for the fourth period, now we're into the 4 s 1 3d for potassium here. And what we notice when we get to the third element in 4s2 and the fourth period is 3d that we go 4 s 2 and then we're back to the 3 d's.

对于第四周期到现在我们来到钾的1,然后我们返回到,我们注意到当我们看到第三个元素,第四周期我们来到,然后我们返回到。

麻省理工公开课 - 化学原理课程节选

So we want to look at any element that has a 3 p orbital filled, but that does not then go on and have a 4 s, because if it had the 4 s filled then we would actually see six lines in the spectrum.

所以，我们要找一找有哪些元素的,3，p，轨道被占据，但没有，4，s，轨道被占据,因为如果，4，s，轨道也被占据了,那我们会在光谱中看到第六条谱线。

麻省理工公开课 - 化学原理课程节选