leetCode(6):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}.

常规版本：遍历链表的方式，事实证明，遍历永远不是最好的！

void reorderList(ListNode* head){//递归、循环效果应该差不多ListNode* p=head;if(head==NULL || head->next==NULL)return;ListNode* currentNode=p;ListNode* nextNode=p->next;ListNode* newNextNode=nextNode;while(nextNode && nextNode->next){ListNode* preNewNextNode=currentNode;ListNode* prePreNewNextNode=currentNode;while(newNextNode->next){preNewNextNode=preNewNextNode->next;newNextNode=newNextNode->next;}prePreNewNextNode->next=NULL;preNewNextNode->next=NULL;currentNode->next=newNextNode;newNextNode->next=nextNode;currentNode=nextNode;nextNode=currentNode->next;newNextNode=nextNode;}}

反转子链表版本：用两个指针，一快一慢遍历整个链表，直到较快的链表到达末尾，慢指针刚好到达中点，然后断开链表。把后段链表反转，再逐个插入。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* reverseList(ListNode* head)    {    if(!head)    return head;    ListNode* p=head;    ListNode* q=p->next;    if(!q)    return head;    ListNode* k=q->next;    p->next=NULL;    while(k)    {    q->next=p;    p=q;    q=k;    k=k->next;    }    q->next=p;    return q;//返回头结点    }    void reorderList(ListNode* head) {    if(head==NULL || head->next==NULL)    return;    ListNode* p=head;    ListNode* fast=p;    ListNode* slow=p;    while(fast && fast->next)    {    fast=fast->next->next;    slow=slow->next;    }    ListNode* subHead=slow->next;    slow->next=NULL;    subHead=reverseList(subHead);    ListNode* p2=subHead;    while(p2!=NULL)    {    ListNode* tmp1=p->next;    ListNode* tmp2=p2->next;    p->next=p2;    p2->next=tmp1;    p2=tmp2;    p=tmp1;    }    }};