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

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

把后面的部分插入前面。

中心思想，将链表重中间断开，然后把后面的链表部分倒序，之后把后面链表的头节点依次插入到head链表里面。

其代码如下。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* reverseList(struct ListNode* head);int LengthList(struct ListNode* head);void reorderList(struct ListNode* head){if(head==NULL||head->next==NULL)return ;int mid;    int length;    struct ListNode* pCurrent=head;length=LengthList(pCurrent);    mid=(length+1)/2; //3 应该是第2个，以后也是在后面一个    //找到中间节点，分成两条链表,//mid 后面哪一个开始struct ListNode* pCurrentmid=head;struct ListNode* pCurrentmidFront;while(mid--){  pCurrentmidFront=pCurrentmid;      pCurrentmid=pCurrentmid->next;  //mid=1，最后一个设置为null}struct ListNode* pCurrentMidafter=pCurrentmid; //3--->1    pCurrentmidFront->next=NULL;  //让他变成两个链表//翻转pCurrentMidafterpCurrentMidafter=reverseList(pCurrentMidafter);//DisplayList(pCurrentMidafter);//DisplayList(head);// //将翻转后的链表插入到head链表中..pCurrent=head;while(pCurrent!=NULL&&pCurrentMidafter!=NULL){        pCurrentmidFront=pCurrentMidafter;       pCurrentMidafter=pCurrentMidafter->next;   //插入  pCurrentmidFront->next=pCurrent->next;      pCurrent->next=pCurrentmidFront;  //放回原位      pCurrent=pCurrentmidFront->next;}}int LengthList(struct ListNode* head) {if(head==NULL)return 0;int length=0;struct ListNode* pCurrent=head;while(pCurrent!=NULL){length++;pCurrent=pCurrent->next;}//free(pCurrent);return length;}struct ListNode* reverseList(struct ListNode* head) {struct ListNode* newhead;newhead=NULL;//这样就不需要翻转了if (head==NULL||head->next==NULL){return head;}struct ListNode* pCurrent=head;  //一样的点while(pCurrent!=NULL){struct ListNode* tmp=pCurrent;pCurrent=pCurrent->next;tmp->next=newhead;   //把节点放在节点前面,插入到链表的最前端newhead=tmp;}return newhead;}