LeetCode题解之Reorder List

1、题目描述

2、题目分析

首先将链表分为两段，然后将后面的一段反转，再合并两个链表。

3、代码

 void reorderList(ListNode* head) {
         if (head == nullptr || head->next == nullptr || head->next->next == nullptr)
             return ;

         ListNode *newH = splitList(head);
         ListNode *rp = reverseList(newH);
         mergeList(head, rp);
     }

    ListNode *reverseList(ListNode* list)
     {
         if (list == NULL || list->next == NULL)
             return list;

         ListNode *p = list;
         ListNode *pn = list->next;
         list->next = NULL;

         while (pn != NULL) {
             ListNode *tmp = pn->next;
             pn->next = p;
             p = pn;
             pn = tmp;
         }
         return p;
     }

     void mergeList(ListNode *head_1, ListNode *head_2)
     {
         ListNode *p;
         ListNode *pm = head_1;
         ListNode *pn = head_2;
         while (pm != NULL && pn != NULL) {
             ListNode *tmpm = pm->next;
             ListNode *tmpn = pn->next;
             pm->next = pn;
             if (tmpm != NULL)
                 pn->next = tmpm;
             pm = tmpm;
             pn = tmpn;
         }
     }

     ListNode *splitList(ListNode *head)
     {
         if (head == NULL || head->next == NULL)
             return head;
         ListNode *pm = head;
         ListNode *pn = head;

         while (pn != NULL && pn->next != NULL) {
             pm = pm->next;
             pn = pn->next->next;

         }

         ListNode *newH = NULL;
         if (pn == NULL) {
             newH = pm;
             ListNode *pt = head;
             while (pt->next != newH) {
                 pt = pt->next;
             }
             pt->next = NULL;
         } else if (pn->next == NULL) {
             newH = pm->next;
             pm->next = NULL;
         }

         return newH;
     }