C++ 程序，用于在链表中从头到尾的第 K 个节点交换第 K 个节点
给定一个单链表，从开始的第 k 个节点与从结束的第 k 个节点交换。不允许交换数据，只能更改指针。在链表数据部分很大的许多情况下（例如学生详细信息行 Name、RollNo、Address 等），此要求可能是合乎逻辑的。指针始终是固定的（大多数编译器为 4 个字节）。
例子：
Input: 1 -> 2 -> 3 -> 4 -> 5, K = 2
Output: 1 -> 4 -> 3 -> 2 -> 5
Explanation: The 2nd node from 1st is 2 and
2nd node from last is 4, so swap them.
Input: 1 -> 2 -> 3 -> 4 -> 5, K = 5
Output: 5 -> 2 -> 3 -> 4 -> 1
Explanation: The 5th node from 1st is 5 and
5th node from last is 1, so swap them.插图：
方法：想法很简单，从头开始找到第k个节点，从最后开始的第k个节点是从头开始的第n-k+1个节点。交换两个节点。
但是有一些极端情况，必须处理
- Y 在 X 旁边
- X 在 Y 旁边
- X 和 Y 相同
- X 和 Y 不存在（k 大于链表中的节点数）
下面是上述方法的实现。
C++
// A C++ program to swap Kth node
// from beginning with kth node from end
#include
using namespace std;
// A Linked List node
struct Node
{
int data;
struct Node* next;
};
/* Utility function to insert
a node at the beginning */
void push(struct Node** head_ref,
int new_data)
{
struct Node* new_node =
(struct Node*)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
/* Utility function for displaying
linked list */
void printList(struct Node* node)
{
while (node != NULL)
{
cout << node->data << " ";
node = node->next;
}
cout << endl;
}
/* Utility function for calculating
length of linked list */
int countNodes(struct Node* s)
{
int count = 0;
while (s != NULL)
{
count++;
s = s->next;
}
return count;
}
/* Function for swapping kth nodes
from both ends of linked list */
void swapKth(struct Node** head_ref, int k)
{
// Count nodes in linked list
int n = countNodes(*head_ref);
// Check if k is valid
if (n < k)
return;
// If x (kth node from start) and
// y(kth node from end) are same
if (2 * k - 1 == n)
return;
// Find the kth node from the beginning
// of the linked list. We also find
// previous of kth node because we
// need to update next pointer of
// the previous.
Node* x = *head_ref;
Node* x_prev = NULL;
for (int i = 1; i < k; i++)
{
x_prev = x;
x = x->next;
}
// Similarly, find the kth node from
// end and its previous. kth node
// from end is (n-k+1)th node from
// beginning
Node* y = *head_ref;
Node* y_prev = NULL;
for (int i = 1; i < n - k + 1; i++)
{
y_prev = y;
y = y->next;
}
// If x_prev exists, then new next of
// it will be y. Consider the case
// when y->next is x, in this case,
// x_prev and y are same. So the statement
// "x_prev->next = y" creates a self loop.
// This self loop will be broken
// when we change y->next.
if (x_prev)
x_prev->next = y;
// Same thing applies to y_prev
if (y_prev)
y_prev->next = x;
// Swap next pointers of x and y.
// These statements also break self
// loop if x->next is y or y->next is x
Node* temp = x->next;
x->next = y->next;
y->next = temp;
// Change head pointers when k is 1 or n
if (k == 1)
*head_ref = y;
if (k == n)
*head_ref = x;
}
// Driver code
int main()
{
// Let us create the following
// linked list for testing
// 1->2->3->4->5->6->7->8
struct Node* head = NULL;
for (int i = 8; i >= 1; i--)
push(&head, i);
cout << "Original Linked List: ";
printList(head);
for (int k = 1; k < 9; k++)
{
swapKth(&head, k);
cout <<
"Modified List for k = " << k << endl;
printList(head);
}
return 0;
} 输出：
Original Linked List: 1 2 3 4 5 6 7 8
Modified List for k = 1
8 2 3 4 5 6 7 1
Modified List for k = 2
8 7 3 4 5 6 2 1
Modified List for k = 3
8 7 6 4 5 3 2 1
Modified List for k = 4
8 7 6 5 4 3 2 1
Modified List for k = 5
8 7 6 4 5 3 2 1
Modified List for k = 6
8 7 3 4 5 6 2 1
Modified List for k = 7
8 2 3 4 5 6 7 1
Modified List for k = 8
1 2 3 4 5 6 7 8复杂性分析：
- 时间复杂度： O(n)，其中 n 是列表的长度。
需要遍历一次列表。 - 辅助空间： O(1)。
不需要额外的空间。
有关详细信息，请参阅完整的文章在链接列表中从头到尾的第 K 个节点交换 Kth 节点！