如何仅在n次打印多页图中的元素

My multimap has an int as a key and a string as the value. The key is the number of times a word has occurred and the value is the word itself. I figured that if I loop through the multimap with my iterator at rbegin and rend, I can just loop through it ten times rather than until it reaches the end to find the top ten words. Any idea how to do this?

我的multimap有一个int作为键，一个字符串作为值。关键是单词出现的次数，值是单词本身。我想如果我在rbegin和rend中使用我的迭代器遍历multimap，我可以循环遍历它十次而不是直到它到达结尾才能找到前十个单词。知道怎么做吗？

2 个解决方案

#1

Sure! Here's one option:

当然！这是一个选项：

unsigned numTimes = 0;
for (auto itr = myMultiMap.rbegin();
     itr != myMultiMap.rend() && numTimes < kMaxTimes;
     ++itr, ++numTimes) {

    /* Do something with itr */

}

#2

Another option would be to use std::advance to find the tenth item (starting from rbegin). With that, you get a valid range, and you can process it with the usual algorithms and such:

另一个选择是使用std :: advance来找到第十个项目（从rbegin开始）。有了它，您将获得一个有效的范围，您可以使用通常的算法处理它，例如：

std::map<int, std::string> word_freqs;

auto first = word_freqs.rbegin();
auto last = std::next(first, 10);

Now, we can (for example) print out those 10 most common words:

现在，我们可以（例如）打印出这10个最常见的单词：

typedef std::pair<int, std::string> T;

std::ostream &operator<<(std::ostream &os, T const &t) { 
    return os << t.second << ": " << t.first;
}

std::copy(first, last, std::ostream_iterator<T>(std::cout, "\n"));

Of course, for robustness, you'd probably want to include a check that the multimap has at least 10 items first (and probably just use rbegin() and rend() if it's smaller than that).

当然，为了实现稳健性，您可能希望包含一个检查，即multimap首先至少有10个项目（如果它小于那个，可能只使用rbegin（）和rend（））。

#1