C++ 返回数组中最小元素的索引
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原文地址: http://stackoverflow.com/questions/13558081/
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Return index of smallest element in array
提问by gosutag
I'm trying to return the index with the smallest element in an array of integers. Am I missing something? After I put my integers in, it doesn't return the index.
我试图返回整数数组中最小元素的索引。我错过了什么吗？在我放入整数后，它不会返回索引。
UPDATE: I get an error at the end of int main()about the array stack being corrupted. Thank you. My code is as follows:
更新：我在结束时收到int main()有关数组堆栈损坏的错误。谢谢你。我的代码如下：
#include <iostream>
#include <conio.h>
using namespace std;
int indexofSmallestElement(double array[], int size);
int main()
{
int size = 10;
double array[10];
for (int i = 0; i <= size; i++)
{
cout << "Enter an integer: " << endl;
cin >> array[i];
}
indexofSmallestElement(array, size);
}
int indexofSmallestElement(double array[], int size)
{
int index = 0;
if (size != 1)
{
int n = array[0];
for (int i = 1; i < size; i++)
{
if (array[i] < n)
{
n = array[i];
index = i;
}
}
}
return index;
}
回答by Nik Bougalis
A number of people have shown you their variant of indexofSmallestElement. I will include mine along with an explanation of whyI think it's better:
许多人向您展示了他们的indexofSmallestElement. 我将包括我的以及为什么我认为它更好的解释：
int indexofSmallestElement(double array[], int size)
{
int index = 0;
for(int i = 1; i < size; i++)
{
if(array[i] < array[index])
index = i;
}
return index;
}
You will notice that I do away with the nvariable, which you used to hold the smallest value encountered so far. I did this because, in my experience, having to keep two different things synchronized can be a source of subtle bugs. Even in this simple case, it was the source of multiplebugs, not the least of which was that your nwas declared an intbut you were assigning values of type doublein it.
您会注意到我取消了n变量，您过去使用该变量保存迄今为止遇到的最小值。我这样做是因为，根据我的经验，必须保持两个不同的东西同步可能会导致微妙的错误。即使在这个简单的例子中，它也是多个错误的根源，其中最重要的是你n被声明为 anint但你在其中分配了类型的值double。
Bottom line: do away with the nvariable and just keep track of one thing: the index.
底线：去掉n变量，只跟踪一件事：索引。
Update: of course, it goes without saying that with C++17 the onlytruly acceptable answer for this question is to use std::min_element:
更新：当然，不用说，对于 C++17 ，这个问题唯一真正可以接受的答案是使用std::min_element：
#include <iostream>
#include <algorithm>
#include <iterator>
#include <cstdint>
#include <array>
#include <vector>
template <typename Iter>
auto indexofSmallestElement(Iter first, Iter last)
{
return std::distance(first,
std::min_element(first, last));
}
template <typename T, std::size_t N>
auto indexofSmallestElement(std::array<T, N> arr)
{
return std::distance(std::begin(arr),
std::min_element(std::begin(arr), std::end(arr)));
}
template <typename T>
auto indexofSmallestElement(std::vector<T> vec)
{
return std::distance(std::begin(vec),
std::min_element(std::begin(vec), std::end(vec)));
}
int main(int, char **)
{
int arr[10] = { 7, 3, 4, 2, 0, 1, 9, 5, 6, 8 };
auto x = indexofSmallestElement(std::begin(arr), std::end(arr));
std::cout
<< "The smallest element in 'arr' is at index "
<< x << ": " << arr[x] << "\n";
std::array<float, 5> fa { 0.0, 2.1, -1.7, 3.3, -4.2 };
auto y = indexofSmallestElement(fa);
std::cout
<< "The smallest element in 'fa' is at index "
<< y << ": " << fa[y] << "\n";
return 0;
}
回答by Coding Mash
It should be n = array[0]instead of array[0] = n. It means you are supposing first element of the array to be the smallest in the beginning and then comparing it with others.
它应该n = array[0]代替array[0] = n. 这意味着您假设数组的第一个元素在开始时是最小的，然后将其与其他元素进行比较。
Moreover in your loop, you are exceeding the bound of your array. The for loop should run till i < sizeand not i <= size.
此外，在您的循环中，您超出了数组的范围。for 循环应该运行 untili < size而不是i <= size。
Your code should be like this..
你的代码应该是这样的..
int indexofSmallestElement(double array[], int size)
{
int index = 0 ;
double n = array[0] ;
for (int i = 1; i < size; ++i)
{
if (array[i] < n)
{
n = array[i] ;
index = i ;
}
}
return index;
}
回答by ashcliffe
inside the loop use array[i] and index = i. Size is the boundary :)
在循环内部使用 array[i] 和 index = i。大小是边界:)
