reorder-list

题目描述

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}.

IDEA

1.先找到链表中间点，由此分开两个链表

2.将后边的链表翻转

3.然后将后边链表的每个节点插入第一个链表，进行合并

CODE

/**
* 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)
return;
ListNode* mid=findMid(head);
ListNode* head1=mid->next;
mid->next=NULL;
reverseList(head1);
merge(head,head1);
}
ListNode* findMid(ListNode* head){
ListNode *fast=head,*slow=head;
while(fast->next&&fast->next->next){
slow=slow->next;
fast=fast->next->next;
}
return slow;
}
void reverseList(ListNode* &head){
if(head==NULL)
return;
ListNode* p=head->next;
ListNode* pre=head;
ListNode* temp=NULL;
while(p){
pre->next = p->next;
temp=p->next;
p->next=head;
head=p;
p=temp;
}
}
void merge(ListNode* head,ListNode* head1){
ListNode *p1=head,*p2=head1,*temp = NULL;
while(p2){
temp=p2->next;
p2->next=p1->next;
p1->next=p2;
p2=temp;
p1=p1->next->next;
}
}
};