c++ - 从二维 vector 中找出最小 vector 元素的更好方法

Question

我正在尝试找出最小尺寸的 vector<int>来自 vector<vector<int>> 的元素.

这是我的有效代码，但它对 vector 进行了两次迭代。

#include <iostream>
#include <vector>
#include <limits>

int main()
{
    std::vector<std::vector<int>> foo = {{1,2,3,4}, {1,2}, {1,2,3,4,5}, {1,2,3}};
    size_t smallestNumElems = std::numeric_limits<size_t>::max();
    for (size_t i = 0; i < foo.size(); ++i)
    {
        const size_t numElems = foo[i].size();
        if (smallestNumElems > numElems)
            smallestNumElems = numElems;
    }
    for (size_t i = 0; i < foo.size(); ++i)
    {
        if (smallestNumElems == foo[i].size())
        {
            for (size_t j = 0; j < foo[i].size(); ++j)
                std::cout << foo[i][j] << '\n';
            break;
        }
    }
}

结果:

1
2
Program ended with exit code: 0

是否有更好的方法来获得相同的结果？

Answer 1

选项 1:

int main()
{
    std::vector<std::vector<int>> foo = {{1,2,3,4}, {1,2}, {1,2,3,4,5}, {1,2,3}};
    size_t smallestNumElems = std::numeric_limits<size_t>::max();
    std::vector<int>* smallestEntry = nullptr; // Store a reference to the smallest entry in here
    for (size_t i = 0; i < foo.size(); ++i)
    {
        const size_t numElems = foo[i].size();
        if (smallestNumElems > numElems) {
            smallestNumElems = numElems;
            smallestEntry = &foo[i];
        }
    }
    for (size_t i = 0; i < smallestEntry->size(); ++i)
       std::cout << smallestEntry->at(i) << '\n';
}

选项 2:

#include <algorithm>
// ... Stuff
int main()
{
    std::vector<std::vector<int>> foo = {{1,2,3,4}, {1,2}, {1,2,3,4,5}, {1,2,3}};

    auto& smallest = *std::min_element(foo.begin(), foo.end(), 
        [](std::vector<int> const& a, std::vector<int> const& b) { // <- could replace std::vector<int> with auto
            return a.size() < b.size(); 
        }
    );

    for (size_t i = 0; i < smallest.size(); ++i)
       std::cout << smallest.at(i) << '\n';
}