#include <algorithm>
#include <iostream>
int main()
{
std::vector<double> v {1.0, 2.0, 3.0, 4.0, 5.0, 1.0, 2.0, 3.0, 4.0, 5.0};
std::vector<double>::iterator biggest = std::max_element(std::begin(v), std::end(v));
//or std::vector<double>::iterator biggest = std::max_element(v.begin(), v.end);
std::cout << "Max element is " << *biggest<< " at position " <<std::distance(std::begin(v), biggest) << std::endl;
//另一方面，取最大位置也可以这样来写：
//int nPos = (int)(std::max_element(v.begin(), v.end()) - (v.begin());
//效果和采用distance(...)函数效果一致
//说明：max_element(v.begin(), v.end()) 返回的是vector<double>::iterator,
//相当于指针的位置，减去初始指针的位置结果即为最大值得索引。
auto smallest = std::min_element(std::begin(v), std::end(v));
std::cout << "min element is " << *smallest<< " at position " <<std::distance(std::begin(v), smallest) << std::endl;
}
输出：
Max element is 5 at position 4
min element is 1 at position 0
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版权声明：本文为CSDN博主「leonardohaig」的原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/leonardohaig/article/details/81484372
