在按降序排序的向量中找到严格小于键的第一个元素 [英] Find the first element strictly less than a key in a vector sorted in descending order
问题描述
我知道可以使用find_if()STL-Algorithm函数来完成此任务,如下所示:
I understand that this task can be accomplished using the find_if() STL-Algorithm function as follows:
long long int k; //k = key
scanf("%lld",&k);
auto it = find_if(begin(v),end(v),[k](auto e){return e<k;});
但是我要求在对数时间内获得结果。由于向量已经按照降序排序,因此我想使用二进制搜索方法。
However I require the result to be obtained in logarithmic time. Since the vector is already sorted in descending order I'd like to use a binary search approach.
我了解STL算法函数 lower_bound
和 upper_bound
保证了对数复杂性。但是,我无法弄清楚如何使用这些函数来获得小于键的第一个元素,而不是大于或等于键的第一个元素。
I understand the STL Algorithm function lower_bound
and upper_bound
guarantee a logarithmic complexity. However I'm unable to figure out how to use these functions to obtain the first element less than a key as opposed to the first element greater than or equal to a key.
例如:
假设我的向量内容为: 21 9 8 7 6 4
Suppose my vector contents are: 21 9 8 7 6 4
我的关键是: 10
我希望输出为 9
,因为它在向量的从左到右扫描中的第一个元素小于 10
。
I would want the output to be 9
, since its the first element in a left to right scan of the vector that is less than 10
.
在这方面的任何帮助都将非常有帮助!
Any help in this regard would be very helpful!
谢谢
推荐答案
您可以对功能对象 std :: greater使用标准算法
。 std :: upper_bound
You can use the standard algorithm std::upper_bound
with the functional object std::greater
.
下面是一个示例。
#include <iostream>
#include <iterator>
#include <functional>
#include <algorithm>
int main()
{
int a[] = { 21, 9, 8, 7, 6, 4 };
int key = 10;
auto it = std::upper_bound(std::begin(a), std::end(a),
key, std::greater<int>());
if (it != std::end(a)) std::cout << *it << std::endl;
}
程序输出为
9
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