leetcode 25. K 个一组翻转链表

给你一个链表，每 k 个节点一组进行翻转，请你返回翻转后的链表。

k 是一个正整数，它的值小于或等于链表的长度。

如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。

示例：

给你这个链表：1->2->3->4->5

当 k = 2 时，应当返回: 2->1->4->3->5

当 k = 3 时，应当返回: 3->2->1->4->5

说明：

你的算法只能使用常数的额外空间。
你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-nodes-in-k-group
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverse(self, head: ListNode, tail: ListNode) -> ListNode:
        prev = None
        p = head
        while prev != tail:
            nex = p.next
            p.next = prev
            prev = p
            p = nex
        return tail, head

    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        hair = ListNode(0)
        hair.next = head
        pre = hair
        while head:
            tail = pre
            for i in range(k):
                tail = tail.next
                if not tail:
                    return hair.next
            nex = tail.next
            head, tail = self.reverse(head, tail)
            pre.next = head
            tail.next = nex
            pre = tail
            head = tail.next
        return hair.next

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    pair myReverse(ListNode* head, ListNode* tail) {
        ListNode *prev = nullptr;
        ListNode *curNode = head;
        ListNode *nextNode = nullptr;
        while (prev != tail) {
            nextNode = curNode->next;
            curNode->next = prev;
            prev = curNode;
            curNode = nextNode;
        }

        return {tail, head};
    }

    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* hair = new ListNode(0);
        hair->next = head;
        ListNode* pre = hair;
        while (head != nullptr) {
            ListNode *tail = pre;
            for (int i = 0; i < k; i++) {
                tail = tail->next;
                if (tail == nullptr) {
                    return hair->next;
                }
            }
            ListNode *nextNode = tail->next;
            tie(head, tail) = myReverse(head, tail);
            pre->next = head;
            tail->next = nextNode;
            pre = tail;
            head = tail->next;
        }
        return hair->next;
    }
};