力扣算法题—143ReorderList

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

Solution：
　　使用快慢指针，得到后半部分的链表
　　使用栈存储后半部分的链表来替代翻转链表

 class Solution {
 public:
     void reorderList(ListNode* head) {
         if (head == nullptr || head->next == nullptr)return;
         ListNode* slow = head, *fast = head;
         while (fast && fast->next)
         {
             slow = slow->next;
             fast = fast->next->next;
         }
         fast = slow->next;
         slow->next = nullptr;
         stack<ListNode*>s;
         while (fast)
         {
             s.push(fast);
             fast = fast->next;
         }
         slow = head;
         while (!s.empty())
         {
             ListNode* p = slow->next;
             slow->next = s.top();
             s.pop();
             slow->next->next = p;
             slow = p;
         }
         return;
     }
 };