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You'll also notice that this thing goes through the entire list, even if the list is sorted before it gets partway through.
你也能注意到，它始终会遍历列表,甚至列表在排序之前,就是有序的也是这样。
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And because, as long as this collection is finite, this thing is just going to walk through. All right?
这个操作就会遍历完它的，对不对？,那么，如果我给大家看看，例如？
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Alex is going to get all the left wing votes, but there aren't enough left wing votes here, right?
亚历克斯将获得他左边左派的选票,但不是所有的左派选票,对吗
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Suppose the individual right to religious liberty were at stake, then, Alex, you could say, on the Hustings.
假设个人信仰自由的权利，岌岌可危，那么，亚历克斯，你在竞选时可以说。
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I just go down the list selecting the smallest person at a time and then I repeat, repeat, repeat but when we actually did out the math or kind of reason through it, the running time, the asymptotic running time of bub-- of Selection Sort was also what?
只需要遍历列表，每次找出最小的元素,然后重复上述步骤，但从数学角度看,选择排序的时间复杂度,又是多少呢？
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We want one epic cohesive story that we tell the world.
我们想告诉世界的是一个完整的历诗。
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In other words it doesn't just return when it finds one, it's going to run through all of them. All right?
但是它将会继续运行,换句话来说当它找到一个结果时,它不会立刻返回结果它会先遍历所有的内容？
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OK? So you can see that, it's just walking through it, and in fact if I look at a couple of another-- another couple of examples, it's been a long day, again, you can see that property.
恩，其实你能看到,他就是遍历列表，你要有空的话,可以再看很多很多其他的例子,你能够看到这个属性。
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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。
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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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I want to try to sum up what we've learned from this discussion, but, first, let's thank John, Alex, and Julia for a really wonderful job.
我想总结一下从这场讨论中我们所学到的，不过首先谢谢约翰，亚历克斯，朱莉娅,谢谢他们的精彩表现。
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google Imagine you want to search through Google to find a particular page. You want to do it in a second. And you're going to just do it the brute force way, assuming you could even reach all of those pages in that time.
假设你想通过,搜索一个特定的网页,你想在1秒内得到结果,然后你就只管去做了,认为能在1秒内遍历遍所有的网页。
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Well basically, all I'd have to do is go through and find every place I checked the constraint, and change it. To incorporate the new constraint.
比较简单，在最上方，基本上,我做的所有事情就是遍历，整段代码然后找到我对比。
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The better way to think about this is, suppose, rather than starting at the beginning, I just grabbed some spot at random, like this one.
不从数组的头开始遍历，我随即的从数组中抓一个元素，比如这个，我去看看这个元素的值。
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Why is the fundamental right to religious liberty different from the right Alex asserts as a fundamental right to private property and to keep what I earn?
为什么宗教信仰权这一基本权利,不同于亚历克斯主张的,私有财产权，和保有自己财产的基本权利？
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But, Alex, Alex, let me make a small case for democracy.
不过，亚历克斯，我举个民主的小案例。
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If it was an unordered list, we were basically stuck with linear search. Got to walk through the whole list to see if the thing is there.
如果是一个未排序的列表,基本上我们就只能使用线性搜索了,通过遍历整个列表来查看。
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Now you might say, wait a minute. Thing's ordered, if I stop part way through and I throw away half the list, doesn't that help me? And the answer is yes, but it doesn't change the complexity.
如果我在半路上停下来,然后不去遍历剩下的数组了,这会有帮助么？答案是有帮助，但这没法改变算法的复杂度，因为我们之前怎么说来着？
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So what would I want to do? I'd like to somehow walk down each of the digits one at a time and add them up. Ah, that's a looping mechanism, right? I need to have some way of walking through them.
去取这个数的,每个数字然后把他们加起来,啊，这是个循环机制对不对？,我得找到一个遍历它们的方法,一种简单的方法可能。
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I'm walking along the list once, taking two things and saying, make sure the biggest one is next.
我遍历一次列表，每次取两个值,确认最大的元素在后面一个。
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So in the first case, I didn't have to do any swaps because 1 was the smallest thing.
所以在第一次遍历中，我没有做,任何的交换因为1就是最小的。
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Suppose I want to find all the divisors of some integer, I want to figure out what all the divisors are that go evenly into it.
也就是遍历所有的整数来寻找,一个数的平方根，让我们再来看一个例子吧,为了找出另外一种的解决方法。
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Figure out what I'm trying to walk through What's the collection of things I'm trying to walk through. Figure out what I want to do at each stage. Figure out what the end test is.
弄清楚需要遍历的东西,弄清楚我需要遍历的东西的集合,弄清楚我在每一步该做什么,弄清楚终结测试。
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I, somehow if, I want to walk through some sequence of data structures, gathering up or doing the same thing, adding ages in until I get a total age and then divide by the number faculty.
我想，以某种方法，来遍历,一些数据结构，把它们相同的属性加到一起，就是一直的把年龄加到一起一直到,得到了一个年龄总数,然后除以员工的数目。
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And that you might look at, for example, that first example, and say, man by this stage it was already sorted, yet it had to go through and check that the third element was in the right place, and then the fourth and then the fifth and then the six.
你可能会去看看例如第一个例子，然后要抱怨，到这里,它已经是排好序的了,但他还是得去遍历查看,第三个元素是不是在正确的位置,然后第四个，第五个，第六个。
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Well let's see. My fall back is, I could just do linear search, walk down the list one at a time, just comparing those things. OK. So that's sort of my base. But what if I wanted, you know, how do I want to get to that sorted list? All right?
我只能做线性搜索了,一次遍历一遍列表，一个一个比较，但如果我想要,那怎样得到有序的列表呢？,现在的一个问题是，我们排序之前？
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Absolutely. So Alex here, Alex ends up with a payoff of what?
完全正确,所以在这亚历克斯,亚历克斯最终的收益是什么
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Because Alex's best response to her running is to not run.
因为亚历克斯对她参选的最佳对策,是不参选
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Think about this compared to a linear search.
线性查找总是从开头开始遍历。
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Alex Harris, who's been a star on the web blog.
亚历克斯·哈里斯，博客上的明星。
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