At that stage we can merge these, and then take this down, do the division merge and bring them back up.
在那里做一次分解,做到这步时,我们可以把这些进行合并,然后把这个拿下来,做分解合并过程后再把它们拿回去。
And the third thing I need to decide is how do I combine? You know, point out to you in the binary search case, combination was trivial. The answer to the final search was just the answer all the way up.
第三个问题是我需要决定如何进行合并?,就你们所知的,在二分查找中所打印出来的,合并的过程是非常简单的,最后查询的结果,就是一路上来所以的结果。
Now, I have to do the merging so I do pointer, pointer.
下面进行合并,同样用指针。
The only big change came in 1960 when they consolidated the ones around Paris, because of the growth of the Paris region, as you can see there.
仅在1960年由于巴黎区的发展,发生了一次改变,他们将巴黎周边的省进行合并,在这里可以看到
So things that are good candidates for divide And conquer are problems where it's easy to figure out how to divide down, and the combination is of little complexity.
因为适合用分治算法解决的问题,最好是能够简单的将问题进行分解,并且合并的过程不是非常的复杂,只要比线性方案要小。
In which I said, you know, I didn't like the fact that things like plus are overloaded, because you can use plus to add strings, you can use plus to add numbers, you can use plus to add floats.
那个玩笑是这样的,你们懂得,我并不喜欢,加法被重载这样的事情,但是你可以用加法,来对字符串进行合并,你可以用加法来加数字。
How much work do I have to do to actually merge them together?
我需要耗费多大的工作量来他们进行合并?,让我来举个例子?
Merge up here. There's a little more code there.
在这里进行合并,这里的代码有点多。
So this is easy. What do I do?
所以这很简单,我要做什么?我进行合并?
Well I need to merge these two lists.
我需要将这两个序列进行合并。
I said it poorly. What's the point?
那么我就可以将两个列表进行合并了?
How hard is it to merge them?
将它们进行合并需要耗费多少时间?嗯?
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