Leetcode - 143. Reorder List

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.
You must do this in-place without altering the nodes’ values.

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:

/*** 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:void reorderList(ListNode* head) {if(head == nullptr)return;ListNode* mid = head;ListNode* p = head;while(p && p->next){mid = mid->next;p = p->next->next;}if(mid->next){ListNode* pnew = reverse(mid->next); // 逆转后半部分mid->next = nullptr; // 截断两个链表p = head;while(p && pnew)  // 将逆转后的链表插入到前半部分链表{ListNode* tmp = p->next;p->next = pnew;pnew = pnew->next;p->next->next = tmp;p = p->next->next;}}}private:ListNode* reverse(ListNode* head){ListNode dummy{-1, head};ListNode* p = &dummy;ListNode* prev = p->next;ListNode* cur = prev->next;while(prev && cur){prev->next = cur->next;cur->next = p->next;p->next = cur;cur = prev->next;}return dummy.next;}
};

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