用于在链接列表中从头到尾的第 K 个节点交换第 K 个节点的Java程序
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。
Java
// A Java program to swap kth
// node from the beginning with
// kth node from the end
class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
class LinkedList
{
Node head;
/* Utility function to insert
a node at the beginning */
void push(int new_data)
{
Node new_node =
new Node(new_data);
new_node.next = head;
head = new_node;
}
/* Utility function for displaying
linked list */
void printList()
{
Node node = head;
while (node != null)
{
System.out.print(node.data + " ");
node = node.next;
}
System.out.println("");
}
/* Utility function for calculating
length of linked list */
int countNodes()
{
int count = 0;
Node s = head;
while (s != null) {
count++;
s = s.next;
}
return count;
}
/* Function for swapping kth nodes from
both ends of linked list */
void swapKth(int k)
{
// Count nodes in linked list
int n = countNodes();
// 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 beginning of
// linked list. We also find previous
// of kth node because we need to update
// next pointer of the previous.
Node x = head;
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;
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 != null)
x_prev.next = y;
// Same thing applies to y_prev
if (y_prev != null)
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 = y;
if (k == n)
head = x;
}
// Driver code
public static void main(String[] args)
{
LinkedList llist =
new LinkedList();
for (int i = 8; i >= 1; i--)
llist.push(i);
System.out.print(
"Original linked list: ");
llist.printList();
System.out.println("");
for (int i = 1; i < 9; i++)
{
llist.swapKth(i);
System.out.println(
"Modified List for k = " + i);
llist.printList();
System.out.println("");
}
}
}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交换第 K 个节点！