STL 向量:移动向量的所有元素-C/C++问题

STL vector: Moving all elements of a vector(STL 向量:移动向量的所有元素)

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问题描述

我有两个 STL 向量 A 和 B，我想清除 A 的所有元素并移动的所有元素>B 到 A 然后清除 B.简单地说，我想这样做:


I have two STL vectors A and B and I'd like to clear all elements of A and move all elements of B to A and then clear out B. Simply put, I want to do this:
std::vector<MyClass> A;
std::vector<MyClass> B;
....
A = B;
B.clear();

由于 B 可能很长，所以需要 k*O(N) 来完成这个操作，其中 k 是一个常数，而 N 是 max(size_of(A), size_of(B)).我想知道是否有更有效的方法来做到这一点.我能想到的一件事是将A和B定义为指针，然后在恒定时间内复制指针并清除B.>

Since B could be pretty long, it takes k*O(N) to do this operation, where k is a constant, and N is max(size_of(A), size_of(B)). I was wondering if there could be a more efficient way to do so. One thing that I could think of is to define A and B as pointers and then copy pointers in constant time and clear out B.

推荐答案

使用 C++11，就这么简单:

Using C++11, it's as simple as:

A = std::move(B);

现在 A 包含以前由 B 持有的元素，而 B 现在是空的.这避免了复制:内部表示只是从 B 移动到 A，所以这是一个 O(1) 解决方案.

Now A contains the elements that were previously held by B, and B is now empty. This avoids copying: the internal representation is simply moved from B to A, so this is an O(1) solution.

至于 C++03，正如Prætorian 所说，您可以交换向量.std::swap 函数有一个特例，它以 std::vectors 作为参数.这有效地交换了内部表示，因此您最终避免创建它们持有的元素的副本.此函数也适用于 O(1) 复杂度.

As for C++03, as Prætorian states, you could swap the vectors. There is a specialization of the std::swap function, which takes std::vectors as its arguments. This effectively swaps the internal representation, so you end up avoiding creating copies of the elements held by them. This function works in O(1) complexity as well.

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