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If we think about the size of a typical cell - excuse me, now I'm getting confused about nuclei.
它大约是10的,负十四次方米，如果我们考虑。
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It implies that this algorithm is calling itself again and again, and again, and on each time the size of the problem I'm trying to sort is being divided by what?
这就说明此算法会一次又一次地调用自己,每次我要排序的问题规模大小,会除以多少呢？
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Stack of size 4, again I'm going there, so I'm going to do this initially, no I'm not, I'm going to start again.
同样的，移动上面四个圆盘，同样也是。
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And when we talk about size, I'm again just going to say the stipulation we're not talking about an absolute classical concept here, but in general we're going to picture it being much further away from the nucleus as we move up in terms of n.
当我们说到尺寸时，我们只是说――,经典的绝对的概念，而是它大约,离原子核有多远。
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So, I'm now multiplying by the result of calling the size of operator, which we've seen before.
现在我乘以运算符的大小,之前我们看到的。
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It says while the index is smaller than the size - of the list, I'm not at the end of the list and I don't have an answer yet, check.
当然是如果我还没检索到目标数的话,因此我会去看看是不是,这里真的看不情。
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I'm done sorting a list of size 2.
现在我已将这2杯排好序了。
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Now this happens to be what we would call a linear process, because the number of times I go around the loop is directly related to the size of the argument. If I double 2 the argument, I'm going to double the number of times I go around the loop. If I increase it by five, 5 I'm going to increase by five the number of times I go around the loop.
这恰好是我们会成为,线性复杂度程序的一个例子,因为我要执行循环的次数是,和输入的参数的大小直接相关的,如果我将这个参数乘以,那么我就要将进行循环的次数也乘以2了，如果我把参数加上,那么循环的次数也要加上5了。
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Well, int is almost always 32 bits at least so far as we've seen so, in fact, any time I draw a pointer hence force I'm just going to draw a square as well because they are, in fact, the same size usually in memory.
嗯，int几乎总是32位的,至少我们至今所看到的是这样的，每次我用一个正方形表示一个指针,因为实际上他们，在内存中的大小是一样的。
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I'm going to let t of b be the number of steps it takes to solve the problem of size b.
我会设立一个t作为,计算指数为b的时候解决问题需要的步骤数。
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If I'm running at nanosecond speed, 1000 n, the size of the problem, whatever it is, is 1000, and I've got a log algorithm, it takes 10 nanoseconds to complete.
如果这个问题的规模，也就是n，是，如果这个问题是对数级的，这将会占据10纳秒的时间，你一眨眼的时间。
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malloc I'm going to now use this new, fancy function called malloc and I'm going to say x gets the return value of malloc of the size of an int.
我现在使用这个新的，别致的函数,然后我指明x，=，malloc的int型数据的大小,的返回值。
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I've got to test to see if I'm in the base case, and if I'm not, then I need to move a tower of size n minus 1, I need to move a tower of size 1, and I need to move a second-sorry about that a second tower of size n minus 1.
首先我看看我是不是在最基本的情况，如果不是的话，我得先做一个N-1个，圆盘的移动，移动一个圆盘,然后再做一次N-1个圆盘的移动。
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If I want to move this stack here, I'm going to take a stack of size n minus 1 move it to the spare spot, now I can move the base disk over, and then I'm going to move that stack of size n minus 1 to there. That's literally what I did, OK?
如果我想移动这些圆盘，我先把从n-1个圆盘1,移动到多余的柱子上去，这样我就能把最底下的圆盘放到这儿了，然后再把从n-1个圆盘放到这儿来,这就是确切的我怎么做的对不对？
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