Leetcode--Reorder List

Problem Description:

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}

.
分析：按照要求交换链表的元素，第一种方法，直接遍历一遍链表将所有节点指针保存在vector中，然后一头一尾不断重建链表即可。具体代码如下：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void reorderList(ListNode *head) {
ListNode *p=head;
int length,i=1,tag=0;
vector<ListNode* > array;
while(p)
{
array.push_back(p);
p=p->next;
}
p=head;
if(array.size()<3) return;
length=array.size()-1;
while(tag<array.size()/2)
{
p->next=array[length--];
p=p->next;
p->next=array[i++];
p=p->next;
tag++;
}
p->next=NULL;
}
};
第二种方法就是找到链表的中间节点，将链表分为两半，然后将后半部分转置，最后将前后两部分进行合并即可，具体代码如下：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void reorderList(ListNode *head) {
if (!head){return;}
if (head->next==NULL){return;}
ListNode *p=head;
ListNode *q=head;

//find the midddle pointer
while (q->next && q->next->next){
p=p->next;
q=q->next->next;
}

//now p is middle pointer
//reverse p->next to end
q = p->next;
while (q->next){
ListNode* tmp = p->next;
p->next = q->next;
q->next = q->next->next;
p->next->next = tmp;
}

//reorder
q = head;
while (p!=q && p->next){
ListNode* tmp = q->next;
q->next = p->next;
p->next = p->next->next;
q->next->next = tmp;
q=q->next->next;
}
return;
}
};