LeetCode 25. Reverse Nodes in k-Group

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本文介绍了一种算法，该算法可以将链表按K个一组进行反转，并返回修改后的链表。如果节点数不是K的倍数，则剩余的节点保持不变。文章提供了具体的实现代码，包括如何使用虚拟头节点简化边界条件处理。

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode *head, int k) {
        if ((head == NULL) || (k == 1)) {
            return head;
        }
        ListNode dummy(0);
        ListNode *last_tail = &dummy;
        ListNode *cur_head, *cur_tail;
        while(head != NULL) {
            cur_head = head;
            for (int i = 1; ((i < k) && (head != NULL)); ++i) {
                head = head->next;
            }
            if (head == NULL) {
                last_tail->next = cur_head;
                break;
            }
            else if (head->next == NULL) {
                last_tail->next = head;
                reverseList(cur_head, k);
                cur_head->next = NULL;
                break;
            }
            else {
                cur_tail = head;
                head = head->next;
                reverseList(cur_head, k);
                last_tail->next = cur_tail;
                last_tail = cur_head;
            }
        }
        return dummy.next;
    }
private:
    void reverseList(ListNode *head, int k) {
        ListNode *pre = head;
        ListNode *temp;
        head = head->next;
        for (int i = 1; i < k; ++i) {
            temp = head->next;
            head->next = pre;
            pre = head;
            head = temp;
        }
    }
};