A log algorithm typically is one where you cut the size of the problem down by some multiplicative factor.

对数级复杂度的算法就是指,通过一系列常量级步数的操作,可以将问题的规模。

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Linear algorithms tend to be things where, at one pass-through, you reduce the problem by a constant amount by one. If you reduce it by two, 1 it's going to be the same thing.

有问题么？,线性复杂度的算法，当进行了一个,常量级步数的操作的时候，将问题的规模缩小了一个。

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You can actually say big O of 1, big O of 1 being constant time, the same number of steps.

其复杂度为O，表示时间是一个常量，所用的步数是相同的。

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And what are the steps I want to count?

我们想数的步数在哪儿呢？

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So supposed that I give you 16 elements to sort, well, following the logic before, the running time involved in sorting 16 elements is gonna be twice the running time 16 of sorting 8 elements, left half and right half plus 16 - and again, a little sanity check, 16 means-- just the merge steps, right?

现在要对16个元素进行排序，根据之前的逻辑，对16个元素排序,要花的时间是对8个元素排序所花时间的,2倍，分别用于左半部分和右半部分，再加上6,这里16是-,做合并的步数，对吗？

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It says, in either case in general, t of b-- and this is where I'm going to abuse notation a little bit but I can basically bound it by t, 12 steps plus t of b over 2.

我可以用一个,比12+t的数代表,这里有点不准确的地方,具体的步数依赖于奇数偶数，但是你们可以看到在两个case中。

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And if I pull it out one more level, 12k it's 12 plus 12 plus t of b over 8, 12 k because I'll have 12 of those to add up, plus t of b over 2 to the k.

总结一下也就是说，在k步以后，总步数应该是，那这种情况什么时候才能停止呢？,才能到达最基本的情况呢？

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