39：Reorder List

题目：Given a singly linked list L : L0−>L1−>...

−>Ln−1−>Ln, reorder it to: L0−>Ln−>L1−>Ln−1−>L2−>Ln−2−> …

You must do this in-place without altering the nodes’ values.
For example, Given {1,2,3,4}, reorder it to {1,4,2,3}.

解析：将链表从中间断开，然后将后半段链表逆序，最后将前半段链表以及逆过序后的后半段链表合并在一起

解题代码如下：

class Solution {public:        void reorderList(ListNode *head) {                if (!head || !head -> next) return;                //计算链表总长度                int len = 0;                ListNode *cur = head;                while (cur != nullptr) {                        ++len;                        cur = cur -> next;                }                int mid = len % 2 != 0 ? len / 2 + 1 : len / 2;                ListNode* mid_ptr = head;                for (int i = 0; i != mid - 1; ++i)                        mid_ptr = mid_ptr -> next;                // 从中间将原先链表分成两个链表                ListNode* head2 = mid_ptr -> next;                mid_ptr -> next = nullptr;                // 将后半段链表逆序                head2 = reverseList(head2);                //将这两个链表再次合并成一个链表                ListNode *cur1 = head, *cur2 = head2;                while (cur2 != nullptr) {                        ListNode* tmp = cur2 -> next;                        cur2 -> next = cur1 -> next;                        cur1 -> next = cur2;                        cur1  = cur1 -> next -> next;                        cur2 = tmp;                }        }private:        ListNode* reverseList(ListNode* head) {                if (!head) return head;                ListNode dummy{-1};                dummy.next = head;                ListNode* cur = head;                while (cur -> next != nullptr) {                        ListNode* tmp = cur -> next;                        cur -> next = tmp -> next;                        tmp -> next = dummy.next;                        dummy.next = tmp;                }                return head;        }};