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It refers solely to trees, and Vallombrosa is the place "where th' Etrurian shades / high overarch't imbow'r."
这个比喻指的就是树木，地点是瓦隆布罗萨,“在瓦隆布罗萨，厄特鲁利亚的光辉逐渐暗淡“
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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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His adventurous song is one that "with no middle flight intends to soar above th' Aonian Mount, while it pursues things unattempted yet in Prose or Rhyme."
他那具有冒险精神的诗歌,“抛开了中途的旅程，要一飞冲天越过爱奥尼神山，同时追寻使用在未经使用的韵律“
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Light thickens, and the CrowMakes Wing to th' Rookie Wood.
天亮了，乌鸦飞回他们的巢穴。
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Now, these Th-cells, helper T-cells, go on to stimulate B-cells, and it's these B-cells that become the mature antibody producing cells that make quantities of antibody that fill up in your body.
这些辅助性T细胞,进一步刺激B淋巴细胞,从而使这些B淋巴细胞,逐渐发育成成熟的抗体生成细胞,可在体内产生大量抗体
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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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This is the initial part of the simile that describes the leaves: "thick as Autumnal Leaves that strow the Brooks / in Vallombrosa, where th' Etrurian shades / high overarch't imbow'r."
这是描写落叶的比喻的开始部分：,“他们人数众多，稠密得像秋天的繁叶,/在瓦隆布罗萨，厄特鲁利亚的光辉逐渐暗淡“
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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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How long does it take me to find the k'th element? Linear. Because I've got to walk my way down it. OK? So in this case, you have linear access. Oh fudge knuckle.
线性的！因为我得从头,向下走一步步走,所以这里是线性访问，哦，有问题了吧。
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It's now exactly that power from which Milton has been excluded by virtue of his blindness -- light: "Hail holy Light, offspring of Heav'n first-born, / Or of th' Eternal Coeternal beam May I express thee unblam'd?"
而是弥尔顿已经被失明剥夺的一种力量,圣光，你是上帝最初的产物，永恒的光束,我是否可以向你表达我无限的敬意“
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It's looking at the i'th and the i plus first element and it's saying, gee, if the i'th element is bigger than the i'th plus first element, what's the next set of three things doing?
他在观察第i个元素,和第i+1个元素,如果第i个元素大于第i+1个元素，接下来的要做哪三件事情？
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So I'm going to check to see if- really can't read that thing, let me do it this way-- right, I'm going to increase the number of compares, and I'm going to check to say, is the thing I'm looking for at the i'th spot in the list?
让我们这样做吧-好-,我会去增加比较次数的计数器,然后去看看在第i处的元素,是不是我要找的呢？,这里s的意思就是,给定数组的？
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So I'm not going to worry whether this item is a vase or a watch or painting I'm just going to say it's the n'th element.
所以我不用担心这个物品,是花瓶还是手表或者油画,我只要说这是第n个元素就好。
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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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Just swapping them, right? I temporarily hold on to what's in the i'th element so I can move the i plus first one in, and then replace that with the i'th element.
交换他们，对么？,临时的保存下第i个元素,然后把第i+1个元素移进来，把i+1的位置替换为第i个元素。
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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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Where n'th is somewhere between and 2 in this case.
而在这个例子中第n个元素,就是2和5之间的一个数。
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