有效的方法来获取迭代器的索引？ [英] Effective way to get the index of an iterator?

问题描述

我迭代一个向量，需要迭代器当前指向的索引。 AFAIK这可以通过两种方式完成：

• it - vec.begin（）



• std :: distance（vec.begin
  
  

  这些方法的利弊是什么？

  解决方案

  code> it - vec.begin（）正是由于Naveen给出的相反的原因：所以如果你将向量转换为列表，它不会编译。如果你在每次迭代中这样做，你可以很容易地将一个O（n）算法转换成一个O（n ^ 2）算法。

  
  
  

  在迭代过程中不要在容器中跳转，将索引保存为第二个循环计数器。

  
  

  I'm iterating over a vector and need the index the iterator is currently pointing at. AFAIK this can be done in two ways:

  • it - vec.begin()
  • std::distance(vec.begin(), it)

  What are the pros and cons of these methods?

  解决方案

  I would prefer it - vec.begin() precisely for the opposite reason given by Naveen: so it wouldn't compile if you change the vector into a list. If you do this during every iteration, you could easily end up turning an O(n) algorithm into an O(n^2) algorithm.

  Another option, if you don't jump around in the container during iteration, would be to keep the index as a second loop counter.

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