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So I guess I should go back, and let me do this correctly this way. Again, I can look at test, test and I guess test now if I want to get the element out-- angle bracket or square bracket?
但是这不是我想要的对不对？,我认为我该回到这里，让我改正这个错误，好，那么我还是对元组,那么现在我想从test中取出元素来？
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So wherever the largest element started out in the list, by the time I get through it, it's at the end.
所以无论最大的元素,在列表的什么位置。
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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。
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If I do this, try and get the 10th element of a list that's only eight long. I get what looks like an error, but it's actually throwing an exception.
如果我来这么做,试着去取长度为8的列表的第十个元素,我貌似能得到一个错误。
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If it's there, I'm done, if not, I keep walking down, and I only stop when I get to a place where the element I'm looking for is smaller than the value in the list., in which case I know the rest of this is too big and I can stop.
并且保持遍历,我只在当当前位置的数组元素,大于目标数时停止,这意味着剩下的元素都比目标元素大，但是其他的情况,我还是要遍历完整个数组。
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In the first spot, I'm going to store something that says, here's how far you have to jump to get to the next element.
我们存下下一个,元素的地址偏移量,然后，用之后连续的几个单元。
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And in the next spot, maybe I've got a really long list, and I'm going to say how to jump to get to the next element.
可能我得存放一个很长的列表了,你们也明白如何定位下一个元素,好的，这很棒，这样我有一个。
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But if, at any point, I get to a place in the list where the thing I'm looking for is smaller than the element in the list, I know everything else in the rest of the list has to be bigger than that, I don't have to bother looking anymore.
比当前位置数组的元素要小,我也就知道后面的数肯定,也都比我的目标数要大了，我就不用再继续进行下去了,这意味着目标数不在这个数组中，我就可以退出了。
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I'm accessing a list. How long does it take for me to get the nth element of a list?
我取得数组的第N个元素,需要多长时间呢？
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Well I have to go to the zero-th element, and say OK, gee, to get to the next element, I've got to jump this here. And to get to the next element, I've got to jump to here, and to get to the next element, I've got to jump to here, until I get there.
再回到第0个元素，好的，然后找下一个元素，跳转到这里，然后再下一个元素，跳到了这里,然后再下一个元素,直到我找到那个元素,这样，我有方法来保存变长的元素。
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Here's the problem. How do I get to the nth- er, the k'th element in the list, in this case?
如何找到第n个元素呢-,在这里，如何找到第k个元素呢？
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To get any location in memory, to get to any value of the list, I simply have to say which element do I want to get, I know that these things are stored in a particular size multiply that index by 4, add it to start, and then it's in a constant amount of time I can go to that location and get out the cell.
取得列表中的任何一个值,简单说来，想要取得列表中的任何元素,我知道这些元素存在特定的大小中,把下标乘以4，加到start上,然后定位到内存单元，并取出值就是固定的时间了,好的，如果元素以固定大小存储。
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