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;
    }