Leetcode--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}

.

下面的方法很笨但很好理解

如list1: 1->2->3->4

构造list1的反转链表：list2: 4->3->2->1

count计数链表共有多少个结点

list1的当前节点指向链表list2的当前节点，再把list1和list2的当前节点指向各自链表的后继节点，重复执行count/2次

处理需分奇偶

/**
* 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==NULL||head->next==NULL||head->next->next==NULL)
return ;
int count=0;
ListNode * list2=NULL;
ListNode * list1=head;
while(list1!=NULL)//构建list1的逆序链表list2
{
if(list2==NULL)
{
list2=(ListNode*)malloc(sizeof(ListNode));
list2->val=list1->val;
list2->next=NULL;
count++;
}
else
{
ListNode * temp=(ListNode*)malloc(sizeof(ListNode));
temp->val=list1->val;
temp->next=list2;
list2=temp;
count++;
}
list1=list1->next;
}
list1=head;
if(count%2==0)//节点个数为偶
{
count /=2;
for(int i=1;i<=count;i++)
{
ListNode * temp1=list1->next;
ListNode * temp2=list2->next;
if(i!=count){
list1->next=list2;
list2->next=temp1;
list1=temp1;
list2=temp2;
}
else
{
list1->next=list2;
list2->next=NULL;
}

}
}
else//节点个数为奇数
{
count=(count-1)/2;
for(int i=1;i<=count;i++)
{
ListNode *temp1=list1->next;
ListNode *temp2=list2->next;
list1->next=list2;
list2->next=temp1;
list1=temp1;
list2=temp2;
}
list1->next=NULL;
}

}
};