143. Reorder List [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.

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.

先对半拆分链表，逆序后面的链表，然后merge成一个链表.

为了便于复习，我把三种链表反转的函数都列出来了.

链表反转见LeetCode:206. Reverse Linked List.我整理的相应的解答

class Solution_143_1 {

public:

     void reorderList(ListNode* head) {

         if (head == nullptr || head-> next == nullptr) return;

        ListNode*slow = head, *fast = head, *prev = nullptr;

         while (fast&&fast-> next) {

            prev = slow;

            slow = slow-> next;

            fast = fast-> next-> next;

        }

        prev-> next = nullptr; //cut at middle

        slow= reverse_3(slow);

         //merge

        ListNode*curr = head;

         while (curr-> next) {

            ListNode*tmp = curr-> next;

            curr-> next = slow;

            slow = slow-> next;

            curr-> next-> next = tmp;

            curr = tmp;

        }

        curr-> next = slow;

    }

     //复习一下链表反转

    ListNode* reverse_1(ListNode*head) {

         if (head == nullptr || head-> next == nullptr) return head;

        ListNode*prev = head;

         for (ListNode*curr = prev-> next, *next = curr-> next;

            curr;

            prev = curr, curr = next, next = next ? next-> next : nullptr) {

            curr-> next = prev;

        }

        head-> next = nullptr;

         return prev;    

    }

    ListNode* reverse_2(ListNode*head) {

         if (head == nullptr || head-> next == nullptr) return head;

        ListNode*prev = head;

        ListNode*pcur = prev-> next;

         while (pcur) {

            ListNode*pnext = pcur-> next;

            pcur-> next = prev;

            prev = pcur;

            pcur = pnext;

        }

        head-> next = nullptr;

         return prev;

    }

    ListNode* reverse_3(ListNode* head) {

         if (head == nullptr || head-> next == nullptr) return head;

        ListNode *p = reverse_3(head-> next);

        head-> next-> next = head;

        head-> next = nullptr;

         return p;

    }

};