用于交换链表中的节点而不交换数据的 C++ 程序

给定一个链表和其中的两个键，交换两个给定键的节点。节点应该通过改变链接来交换。当数据包含许多字段时，在许多情况下交换节点的数据可能会很昂贵。

可以假设链表中的所有键都是不同的。

例子：

Input : 10->15->12->13->20->14,  x = 12, y = 20
Output: 10->15->20->13->12->14

Input : 10->15->12->13->20->14,  x = 10, y = 20
Output: 20->15->12->13->10->14

Input : 10->15->12->13->20->14,  x = 12, y = 13
Output: 10->15->13->12->20->14

这可能看起来是一个简单的问题，但却是一个有趣的问题，因为它需要处理以下情况。

1. x 和 y 可能相邻也可能不相邻。
2. x 或 y 都可以是头节点。
3. x 或 y 可能是最后一个节点。
4. x 和/或 y 可能不存在于链表中。

如何编写一个干净的工作代码来处理上述所有可能性。

这个想法是首先在给定的链表中搜索 x 和 y。如果其中任何一个都不存在，则返回。在搜索 x 和 y 时，跟踪当前和以前的指针。首先更改前一个指针的下一个，然后更改当前指针的下一个。

下面是上述方法的实现。

C++

// C++ program to swap the nodes of linked list rather
// than swapping the field from the nodes.
#include 
using namespace std;
 
// A linked list node
class Node
{
    public:
    int data;
    Node* next;
};
 
// Function to swap nodes x and y in
// linked list by changing links
void swapNodes(Node** head_ref,
               int x, int y)
{
    // Nothing to do if x and y
    // are same
    if (x == y)
        return;
 
    // Search for x (keep track of
    // prevX and CurrX
    Node *prevX = NULL,
         *currX = *head_ref;
    while (currX && currX->data != x)
    {
        prevX = currX;
        currX = currX->next;
    }
 
    // Search for y (keep track of
    // prevY and CurrY
    Node *prevY = NULL,
         *currY = *head_ref;
    while (currY && currY->data != y)
    {
        prevY = currY;
        currY = currY->next;
    }
 
    // If either x or y is not present,
    // nothing to do
    if (currX == NULL ||
        currY == NULL)
        return;
 
    // If x is not head of linked list
    if (prevX != NULL)
        prevX->next = currY;
    else
 
        // Else make y as new head
        *head_ref = currY;
 
    // If y is not head of linked list
    if (prevY != NULL)
        prevY->next = currX;
    else // Else make x as new head
        *head_ref = currX;
 
    // Swap next pointers
    Node* temp = currY->next;
    currY->next = currX->next;
    currX->next = temp;
}
 
// Function to add a node at the
// beginning of List
void push(Node** head_ref,
          int new_data)
{
    // Allocate node
    Node* new_node = new Node();
 
    // Put in the data
    new_node->data = new_data;
 
    // Link the old list off the new node
    new_node->next = (*head_ref);
 
    // Move the head to point to the
    // new node
    (*head_ref) = new_node;
}
 
// Function to print nodes in a
// given linked list
void printList(Node* node)
{
    while (node != NULL)
    {
        cout << node->data << " ";
        node = node->next;
    }
}
 
// Driver code
int main()
{
    Node* start = NULL;
 
    /* The constructed linked list is:
       1->2->3->4->5->6->7 */
    push(&start, 7);
    push(&start, 6);
    push(&start, 5);
    push(&start, 4);
    push(&start, 3);
    push(&start, 2);
    push(&start, 1);
 
    cout << "Linked list before calling swapNodes() ";
    printList(start);
    swapNodes(&start, 4, 3);
    cout << "Linked list after calling swapNodes() ";
    printList(start);
    return 0;
}
// This is code is contributed by rathbhupendra

输出：

Linked list before calling swapNodes() 1 2 3 4 5 6 7 
Linked list after calling swapNodes() 1 2 4 3 5 6 7 

时间复杂度： O(n)

辅助空间： O(1)

优化：上面的代码可以优化为在单次遍历中搜索 x 和 y。两个循环用于保持程序简单。

更简单的方法：

C++

// C++ program to swap two given nodes
// of a linked list
#include 
 
using namespace std;
 
// A linked list node class
class Node
{
    public:
    int data;
    class Node* next;
 
    // constructor
    Node(int val, Node* next)
        : data(val)
        , next(next)
    {
    }
 
    // print list from this
    // to last till null
    void printList()
    {
        Node* node = this;
 
        while (node != NULL)
        {
            cout << node->data << " ";
            node = node->next;
        }
 
        cout << endl;
    }
};
 
// Function to add a node
// at the beginning of List
void push(Node** head_ref,
          int new_data)
{
    // Allocate node
    (*head_ref) = new Node(new_data,
                           *head_ref);
}
 
void swap(Node*& a, Node*& b)
{
    Node* temp = a;
    a = b;
    b = temp;
}
 
void swapNodes(Node** head_ref,
               int x, int y)
{
    // Nothing to do if x and
    // y are same
    if (x == y)
        return;
 
    Node **a = NULL, **b = NULL;
 
    // Search for x and y in the linked list
    // and store their pointer in a and b
    while (*head_ref)
    {
        if ((*head_ref)->data == x)
        {
            a = head_ref;
        }
 
        else if ((*head_ref)->data == y)
        {
            b = head_ref;
        }
 
        head_ref = &((*head_ref)->next);
    }
 
    // If we have found both a and b
    // in the linked list swap current
    // pointer and next pointer of these
    if (a && b)
    {
        swap(*a, *b);
        swap(((*a)->next), ((*b)->next));
    }
}
 
// Driver code
int main()
{
    Node* start = NULL;
 
    // The constructed linked list is:
    // 1->2->3->4->5->6->7
    push(&start, 7);
    push(&start, 6);
    push(&start, 5);
    push(&start, 4);
    push(&start, 3);
    push(&start, 2);
    push(&start, 1);
 
    cout << "Linked list before calling swapNodes() ";
    start->printList();
    swapNodes(&start, 6, 1);
    cout << "Linked list after calling swapNodes() ";
    start->printList();
}

输出：

Linked list before calling swapNodes() 1 2 3 4 5 6 7 
Linked list after calling swapNodes() 6 2 3 4 5 1 7 

时间复杂度： O(n)

辅助空间： O(1)

请参考完整的文章在链表中交换节点而不交换数据以获取更多详细信息！