Reorder List leetcode

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}

.
分析：先用快慢指针找到链表的中点，然后翻转链表后半部分，再和前半部分组合。需要注意的是把链表分成两半时，前半段的尾节点要置为NULL，翻转链表时也要把尾节点置为NULL。代码如下
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void reorderList(ListNode *head) {
if(head == nullptr || head->next == nullptr)
return;

ListNode* fast = head;
ListNode* slow = head;
while(fast->next != nullptr && fast->next->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
}
fast = slow->next;
slow->next = nullptr;
fast = reverse(fast);
slow = head;
ListNode* slowNext;
ListNode* fastNext;
while(fast != nullptr) {
slowNext = slow->next;
fastNext = fast->next;
fast->next = slow->next;
slow->next = fast;
slow = slowNext;
fast = fastNext;
}
}
ListNode* reverse(ListNode* head) {
if(head == nullptr || head->next == nullptr)
return head;
ListNode* helper = new ListNode(0);
helper->next = head;
ListNode* pre = helper;
ListNode* last = head;
ListNode* cur = last->next;
while(cur != nullptr) {
last->next = cur->next;
cur->next = pre->next;
pre->next = cur;
cur = last->next;
}
return helper->next;
}
};