使用迭代器获取 std::list 中的下一个元素

Question

我正在尝试获取 std::list 中连续元素之间的差异。我尝试了以下解决方案，但正如this 线程所说，我需要制作迭代器的副本并增加它。不确定这是什么意思，因为向迭代器添加数字也会导致错误。我在这里想念什么。我使用的是以前版本的 C++，但不是 C++ 11

#include "stdafx.h"
#include <list>
#include <iostream>
using namespace std;

int main()
{
    std::list<int> myList;
    //for (int i = 10;i < 15;i++)
    myList.push_back(12);
    myList.push_back(15);
    myList.push_back(18);
    myList.push_back(19);
    myList.push_back(25);

    for (std::list<int>::const_iterator itr = myList.begin();itr != myList.end();++itr)
    {
        int x = *itr;
        int y = *(itr + 1);
        int diff = std::abs(x - y);
        cout << diff << "\n";
    }

    return 0;
}

Answer 1

使用adjacent_difference 怎么样？

std::adjacent_difference(myList.begin(), myList.end(), tempList );

See Here

---

如果您对实现感兴趣，请查看附加链接中Possible Implementation 部分的实现方式。您所要做的就是用列表迭代器替换，输出只显示在屏幕上，或者存储它。

Answer 2

使用增量运算符移动迭代器。

auto x = itr;
auto y = itr;
if(++y!=myList.end())
{
    int diff = std::abs(*x - *y);
    cout << diff << "\n";
}

Answer 3

您可以复制迭代器并运行两个，但我认为在循环之外获取第一个值然后使用循环遍历第二个值更容易，在继续下一个之前用第二个值重置第一个值.

类似这样的：

#include <list>
#include <iostream>

int main()
{
    std::list<int> myList;
    myList.push_back(12);
    myList.push_back(15);
    myList.push_back(18);
    myList.push_back(19);
    myList.push_back(25);
    std::list<int>::const_iterator itr = myList.begin();
    if(itr != myList.end()) // is a first value
    {
        int last = *itr; // cache it
        for (itr++; itr != myList.end(); itr++) // get next value
        { 
            int current = *itr; //cache it
            int diff = std::abs(last - current);
            std::cout << diff << "\n";
            last = current; // update last value
        }
    }
    return 0;
}

Answer 4

您可以使用std::advance() 进入下一个迭代器

int x = *itr;
std::list<int>::const_iterator itr2 = itr;
std::advance(itr2,1);
if(itr2==myList.end())
    break;
int y = *(itr2);
int diff = std::abs(x - y);
cout << diff << "\n";

编辑：如果c++ 11 可用，请参阅std::next

Answer 5

#include <list>
#include <iostream>

int main()
{
    std::list<int> myList;
    //for (int i = 10;i < 15;i++)
    myList.push_back(12);
    myList.push_back(15);
    myList.push_back(18);
    myList.push_back(19);
    myList.push_back(25);
    for (std::list<int>::const_iterator itr = myList.begin();
                              itr != std::prev(myList.end());
                                                      ++itr)
    {
        std::list<int>::const_iterator nextIt = std::next(itr);
        int diff = *itr - *nextIt;
        std::cout << diff << "\n";
    }

    return 0;
}

使用std::next(itr) 获取下一个迭代器 并在 for 循环中使用 std::prev(myList.end()) 来获取列表中最后一个元素的前一个元素。

您还可以更改 for 循环并使用 std::advance() 获取下一个迭代器，而无需使用 std::next 和 std::prev

std::list<int>::const_iterator itr = myList.begin();
while (true)
{
    int prevValue= *itr;
    std::advance(itr, 1);
    if (itr == myList.end())
    {
        break;
    }
    int diff = prevValue - *itr;
    std::cout << diff << "\n";

}

Answer 6

另一种方法是简单地跟踪前一个列表元素的值，而不是迭代器，然后从迭代器当前指向的元素中减去前一个值；像这样：

#include <list>
#include <cmath> // need cmath for std::abs
#include <iostream>
using namespace std;

int main()
{
  std::list<int> myList;
  //for (int i = 10;i < 15;i++)
  myList.push_back(12);
  myList.push_back(15);
  myList.push_back(18);
  myList.push_back(19);
  myList.push_back(25);
  int PreviousElement = 0;
  bool Start = false;
  for (std::list<int>::const_iterator itr = myList.begin(); itr != myList.end(); ++itr)
  {
    if (Start)
    {
      int diff = std::abs(*itr - PreviousElement);
      cout << diff << "\n";
    }
    else
    {
      Start = true;
    }
    PreviousElement = *itr;
  }

  return 0;
}

Answer 7

我认为你需要两个迭代器来解决这个问题 这是我的代码

#include "stdafx.h"
#include <list>

#include <iterator>
#include <iostream>

using namespace std;

int main()
{
    std::list<int> myList;
    //for (int i = 10;i < 15;i++)
    myList.push_back(12);
    myList.push_back(15);
    myList.push_back(18);
    myList.push_back(19);
    myList.push_back(25);
    int x = 0;
    int y = 0;
    std::list<int>::iterator itr;
    std::list<int>::iterator it_next;

    for (itr = myList.begin(), it_next = ++myList.begin();it_next != myList.end();itr++, it_next++)
    {

        x = *itr;
        y = *it_next;
        int diff = abs(x - y);
        cout << diff << "\n";
    }

    system("PAUSE");
    return 0;
}

我在 VC++2005 中构建它，但在 C++ 中应该可以正常工作
希望这会对你有所帮助:)