给定一个XOR链表和一个整数K ，任务是反转给定XOR链表中的每K个节点。
例子：
Input: XLL = 7< – > 6 < – > 8 < – > 11 < – > 3, K = 3
Output: 8 < – > 6 < – > 7 < – > 3 < – > 11
Explanation:
Reversing first K(= 3) nodes modifies the Linked List to 8 < – > 6 < – > 7 < – > 11 < – > 3.
Reversing remaining nodes of the Linked List to 8 < – > 6 < – > 7 < – > 3 < – > 11.
Therefore, the required output is 8 < – > 6 < – > 7 < – > 3 < – > 11.
Input: XLL = 7 < – > 6 < – > 8 < –> 11 < – > 3 < – > 1 < – > 2 < – > 0, K = 3
Output: 8 < – > 6 < – > 7 < – > 1 < – > 3 < – > 11 < – > 0 < – > 2
方法：想法是递归地反转一组中XOR链表中的每个K个节点，并将每组K个节点的第一个节点连接到其前一组节点的最后一个节点。递归函数如下：
RevInGrp(head, K, N)
{
reverse(head, min(K, N))
if (N < K) {
return head
}
head->next = RevGInGrp(next, K, N – K)
}
请按照以下步骤解决问题：
- 反转XOR链表的前K个节点，然后递归地反转大小为K的组中的其余节点。如果剩余节点数少于K ，则只需反转剩余节点数即可。
- 最后，将每个组的第一个节点连接到其前一组的最后一个节点。
下面是上述方法的实现：
C
// C program to implement
// the above approach
#include
#include
#include
// Structure of a node
// in XOR linked list
struct Node {
// Stores data value
// of a node
int data;
// Stores XOR of previous
// pointer and next pointer
struct Node* nxp;
};
// Function to find the XOR of address
// of two nodes
struct Node* XOR(struct Node* a,
struct Node* b)
{
return (struct Node*)((uintptr_t)(a)
^ (uintptr_t)(b));
}
// Function to insert a node with
// given value at given position
struct Node* insert(struct Node** head,
int value)
{
// If XOR linked list is empty
if (*head == NULL) {
// Initialize a new Node
struct Node* node
= (struct Node*)malloc(
sizeof(struct Node));
// Stores data value in
// the node
node->data = value;
// Stores XOR of previous
// and next pointer
node->nxp = XOR(NULL, NULL);
// Update pointer of head node
*head = node;
}
// If the XOR linked list
// is not empty
else {
// Stores the address
// of current node
struct Node* curr = *head;
// Stores the address
// of previous node
struct Node* prev = NULL;
// Initialize a new Node
struct Node* node
= (struct Node*)malloc(
sizeof(struct Node));
// Update curr node address
curr->nxp = XOR(node, XOR(NULL,
curr->nxp));
// Update new node address
node->nxp = XOR(NULL, curr);
// Update head
*head = node;
// Update data value of
// current node
node->data = value;
}
return *head;
}
// Function to print elements of
// the XOR Linked List
void printList(struct Node** head)
{
// Stores XOR pointer
// in current node
struct Node* curr = *head;
// Stores XOR pointer of
// in previous Node
struct Node* prev = NULL;
// Stores XOR pointer of
// in next node
struct Node* next;
// Traverse XOR linked list
while (curr != NULL) {
// Print current node
printf("%d ", curr->data);
// Forward traversal
next = XOR(prev, curr->nxp);
// Update prev
prev = curr;
// Update curr
curr = next;
}
}
// Reverse the linked list in group of K
struct Node* RevInGrp(struct Node** head,
int K, int len)
{
// Stores head node
struct Node* curr = *head;
// If the XOR linked
// list is empty
if (curr == NULL)
return NULL;
// Stores count of nodes
// reversed in current group
int count = 0;
// Stores XOR pointer of
// in previous Node
struct Node* prev = NULL;
// Stores XOR pointer of
// in next node
struct Node* next;
// Reverse nodes in current group
while (count < K && count < len) {
// Forward traversal
next = XOR(prev, curr->nxp);
// Update prev
prev = curr;
// Update curr
curr = next;
// Update count
count++;
}
// Disconnect prev node from the next node
prev->nxp = XOR(NULL, XOR(prev->nxp, curr));
// Disconnect curr from previous node
if (curr != NULL)
curr->nxp = XOR(XOR(curr->nxp, prev), NULL);
// If the count of remaining
// nodes is less than K
if (len < K) {
return prev;
}
else {
// Update len
len -= K;
// Recursively process the next nodes
struct Node* dummy
= RevInGrp(&curr, K, len);
// Connect the head pointer with the prev
(*head)->nxp = XOR(XOR(NULL,
(*head)->nxp),
dummy);
// Connect prev with the head
if (dummy != NULL)
dummy->nxp = XOR(XOR(dummy->nxp, NULL),
*head);
return prev;
}
}
// Driver Code
int main()
{
/* Create following XOR Linked List
head-->7<–>6<–>8<–>11<–>3<–>1<–>2<–>0*/
struct Node* head = NULL;
insert(&head, 0);
insert(&head, 2);
insert(&head, 1);
insert(&head, 3);
insert(&head, 11);
insert(&head, 8);
insert(&head, 6);
insert(&head, 7);
// Function Call
head = RevInGrp(&head, 3, 8);
// Print the reversed list
printList(&head);
return (0);
} 输出：
8 6 7 1 3 11 0 2
时间复杂度： O(N)
辅助空间： O(N / K)