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.
1

Example 2:

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

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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