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;    }};