具有STL的两个序列中的匹配数 [英] number of matches in two sequences with STL

问题描述

好的，这里是另一个问题的如何做到更好的STL？系列。

OK, here is yet another question of the "How to do it better in STL?" series.

我们有两个范围，由first1，last1和first2指定。我们想从[0，last1-first1]找到不同i的数字，使得 *（first1 + i）== *（first2 + i）

We have two ranges, designated by first1, last1, and first2. We want to find the number of different i's from [0, last1-first1] such that *(first1 + i) == *(first2 + i)

例如：

{a, b, c, d, d, b, c, a}
{a, a, b, c, d, c, c, a}
 ^           ^     ^  ^ 

对于这两个范围，答案是4。

For these two ranges the answer is 4.

有很好的STL方法吗？

Is there a good STL way to do it? I mean preferrably without any manual for's, while's etc. Thanks!

推荐答案

std::inner_product(first1, last1, first2, 0, std::plus<T>(), std::equal_to<T>());

UPDATE

Konrad在下面的注释中指出，这依赖于从 bool 到 int 的一些非法的隐式转换。虽然这是完全合法的，但是可以这样避免：

Konrad has pointed out in the comments below that this relies on slightly nefarious implicit conversion from bool to int. Whilst this is completely legal, it can be avoided thus:

template <typename T>
int match(const T &a, const T &b) { return (a == b) ? 1 : 0; }

std::inner_product(first1, last1, first2, 0, std::plus<T>(), match<T>);

这篇关于具有STL的两个序列中的匹配数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！