创建排序向量的索引向量 [英] Creating a vector of indices of a sorted vector
问题描述
变量 x
是 n
整数的向量,我想按升序对向量进行排序。但是,出于超出此问题范围的原因,我希望保持不变。因此,我不是真正地对 x
的内容进行排序,而是要创建另一个 n
索引的向量,其中每个索引如果要对 x
进行排序,则表示 x
中的相应值。
Variable x
is a vector of n
ints, and I want to sort the vector in ascending order. However, for reasons outside the scope of this question, I want to vector to remain untouched. Therefore, rather than actually sorting the contents of x
, I want to create another vector of n
indices, where each index refers to the respective value in x
, if x
were to have been sorted.
例如:
std::vector<int> x = {15, 3, 0, 20};
std::vector<int> y;
// Put the sorted indices of x into the vector y
for (int i = 0; i < 4; i++)
{
std::cout << y[i];
}
应提供输出:
2
1
0
3
对应于x中的值:
0
3
15
20
我能想到很多及时的实现方法,但我想知道STL是否有内置的东西可以为我高效执行此操作?
I can think of plenty of timely ways of implementing this, but I'm wondering whether the STL has something build-in to perform this efficiently for me?
推荐答案
1) y
作为索引的向量(整数范围)
1) Create y
as a vector of index (an integer range)
2)对此进行排序使用比较器返回范围,该比较器使用标准库返回 x
中的索引元素,从而得出:
2) Sort this range with a comparator that returns the indexes elements from x
Using the Standard Library, that gives :
#include <iostream>
#include <vector>
#include <algorithm>
int main() {
std::vector<int> x = {15, 3, 0, 20};
std::vector<int> y;
std::vector<int> y(x.size());
std::size_t n(0);
std::generate(std::begin(y), std::end(y), [&]{ return n++; });
std::sort( std::begin(y),
std::end(y),
[&](int i1, int i2) { return x[i1] < x[i2]; } );
for (auto v : y)
std::cout << v << ' ';
return 0;
}
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