Write a one-pass algorithm with o (1) extra space to determine, if your path crosses itself, or not.
编写一个一趟遍历算法,并且只使用o(1)的额外空间来确定路径是否自相交。
To program slicing, the concept of SDG and a two-pass algorithm are effective to solve "calling-context" problem.
系统依赖图的概念及两阶段图形可达性算法的出现,则有效解决了程序切片的过程调用问题。
This article presents a two-pass algorithm to render the scene including participating media, and this algorithm is an expansion of the rendering to solid objects with photon mapping.
本文依据光子映射对实体物体的渲染,将其扩展到对包含参与介质的场景的渲染,为此提出了一个两路的渲染算法。
Why don't I just keep track on each pass through the algorithm whether I have done any swaps?
我为什么不仅仅跟踪算法的,每一遍来看看我做了哪些交换呢?
All right? Because if I don't do any swaps on a pass through the algorithm, then it says everything's in the right order.
好么?因为如果我,这一遍没有做任何交换,那么就意味着已经排好序了。
And in this case, we go from 8 to 4 to 2 to 1 three times and then on each iteration of this algorithm, each pass across the board I'm touching N numbers, so that means I'm doing N things, log N times.
在这个例子中,我们从8得到4,到2,再到1,是3次,在这个算法的每次迭代中,每一趟我都会操作N个数,也就是所我每次要做N步操作,一共要做,log,N,次。
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