So it's certainly at least linear in the length of the list. For each starting point, what do I do?
它至少是线性的计算列表的长度,每次到了循环开始的点?
I just doubled the indentation each time so you can see it. So each successive call, notice what's happening. The argument is getting reduced. And we're going another level in. When we get down to this point, we're calling it with just a string of length one.
因此每次成功调用注意它的过程,我们的命题不断简化,而且我们不断深入嵌套,当我们走到这一步的时候,我们就是在调用它处理,仅有一个元素的字符串了。
At each point, er is a different vector pointing in the radial direction of length one.
矢量 er 在每一点处都不同,方向都从圆心指向该点,模长为1
It's going to let j run over the length of the list all right, so it's going to start at some point to move down, and then it's going to let i run over range, that's just one smaller, and what's it doing there?
就在这里,会做什么呢?,我们让j遍历这个列表,好的,它要从某一定开始移动,然后让i也跑遍整个范围,就是比之要小一,会发生什么?
er is a vector at each point of length one pointing radially away from the center.
r 是一个模长恒为 1 的矢量,方向沿半径向外
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