# 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}.

这题比较简单,其实就是将链表的左右两边合并,只是合并的时候右半部分需要翻转一下。

主要有三步:

+ 快慢指针找到切分链表
+ 翻转右半部分
+ 依次合并

代码如下:

```c++
class Solution {
public:
void reorderList(ListNode *head) {
if(head == NULL || head->next == NULL) {
return;
}

ListNode* fast = head;
ListNode* slow = head;


//快慢指针切分链表
while(fast->next != NULL && fast->next->next != NULL){
fast = fast->next->next;
slow = slow->next;
}

fast = slow->next;
slow->next = NULL;

//翻转右半部分
ListNode dummy(0);
while(fast) {
ListNode* n = dummy.next;
dummy.next = fast;
ListNode* nn = fast->next;
fast->next = n;
fast = nn;
}

slow = head;
fast = dummy.next;

//依次合并
while(slow) {
if(fast != NULL) {
ListNode* n = slow->next;
slow->next = fast;
ListNode* nn = fast->next;
fast->next = n;
fast = nn;
slow = n;
} else {
break;
}
}
}
};
```