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Right, so s of i saying, given the list, look at the i'th element, true is it the same thing?
第2个元素，看他们是不是相等呢？,如果他们相等，好，把答案设置为？
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And if I'm walking down the list, this is probably order of the length of the list s because I'm looking at each element once.
这可能大概就是数组的长度，因为我会遍历数组中的每个元素一次,现在你可能会想，等等，数组已经排好序了。
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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。
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I look at the rest of the list to decide what's the element to swap into the next place.
我会查看列表剩下的部分,来看看需要交换哪个元素。
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Now, compare the first element in each of these lists. Two is less than three, so two ought to be the next element of the list.
现在，比较每个列表的第一个元素,2要比3小，所以2应该是合并后列表的，下一个元素，然后你们就知道了。
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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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Because if you did what I suggested with the list, the time to look up the key would be linear in the length of the list. You'd have to look at each element until you found the key.
字典是用一种很神奇的，叫做散列法的算法,来实现的，后面我们将,会学到一点关于。
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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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OK. And I'm going to look to see, is a particular element inside of that list, and again, I'll remind you, that's just giving me the integers from zero up to 9999 something or other.
一个又大又长的列表吧，好的,我打算查找列表中的一个特定元素，我会告诉你，我用的是0到大概是,9999左右的整数。
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The easy thing to do would be the following: start at the front end of the list, check the first element. If it's the thing I'm looking for, I'm done. It's there. If not, move on to the next element. And keep doing that.
从数组的第一个元素开始：,如果这是目标元素那么结束,如果不是的话就继续比较下面的元素，并一直这么比较下去,但如果，在某一点我发现我要寻找的数字。
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I've got a list, walk you through it an element at a time, do I look at each element of the list more than once?
你一次只能得到他的一个元素,我是不是把数组里面的每个元素，都过了大于等于一次？,你不这样认为么？大家有什么建议？
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Lisp The problem with that technique, while it's very general, is how long does it take me to find the i'th element of the list?
很多编程语言包括，都是用了这种技术,这个问题就是，当数组很大的时候？
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And that's like kind of hoping that the element you're searching for is the first in the list and the last in the list. Maybe.
也就是希望要找的元素,在数组的头部或者尾部,情况可能是这样的。
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That one's not so obvious. So let's think about this for a second. To sort a list in linear time, would say, I have to look at each element in the list at most a constant number of times.
所以让我们来思考一会,要在线性时间能排序，列表里每个元素最多被使用常数次,不一定是一次，对吧。
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Hmm. Well, wait a minute. If I want to sort a list, I'll take one element, I've got to look at probably a lot of the other elements in the list in order to decide where it goes. And that suggests it's going to depend on how long the list is.
我会取一个元素，然后去查看列表,来决定要把那个元素放到哪里,这就意味着它依赖于列表的长度,对，这没什么说服力，但是事实上。
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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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It took me seven comparisons, because I can take advantage of the fact I know I only ever have to look at the first element of each sub-list. Those are the only things I need to compare, and when I run out of one list, I just add the rest of the list in.
进行了7次对比，因为我可以利用我知道的优势：,每次只需要比较每个子列表的第一个元素,那才是我需要进行对比的内容,当一个列表的元素处理完了,只需要将另一个列表剩下的元素直接添加进去。
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That's accessing an element of a list.
我们需要确认这一点。
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I'll let you just grok it but you can see it's basically doing what I did over there. Setting up two indices for the two sub-list,it's just walking down, finding the smallest element, putting it into a new list. When it gets to the end of one of the lists, it skips to the next part, and only one of these two pieces will get called because only one of them is going to have things leftovers.
你们可以大体的浏览一下,但是它们基本就是我在那里所做的事情,为两个子列表设置了两个指针,指针顺着列表走下去，找到最小的元素,把它放入到一个新的列表中去,当它走到一个列表的尾部时,它会跳到下部分去,两部分中只有一个会被执行,因为只有一个会有元素剩余。
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I've got to count my way down, which means that the access would be linear in the length of the list to find the i'th element of the list, and that's going to increase the complexity.
的位置并去访问，然后继续下去，也就意味着,找到数组中的第i个元素的方法,是关于数组的长度呈线性复杂度的，这回增加算法的复杂度。
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With this, if I can assume that accessing the i'th element of a list is constant, then you can't see that the rest of that analysis looks just like the log analysis I did before, and each step, no matter which branch I'm taking, I'm cutting the problem down in half.
读取数组中的第i个元素,是个常量时间的操作的话，我也就能像以前那样得到,这个算法是对数级复杂度的分析,并且每一步不管我选择哪个区间，我都可以把问题的规模缩小一半。
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You can also assign to a list, or to an element of a list. So ivy sub 1, f -15 or example, could be assigned minus 15, and that will actually mutate the list.
此外你还可以给一个数组，或者数组内部的元素赋值,比如ivy。sub，可以被赋值为，这确实改变了数组。
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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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