算法题目中常考察的链表操作无非以下几种：
- 链表反转
- 链表合并
- 寻找链表中点
- 寻找链表倒数第 K 个节点
- 删除链表节点
- 判断链表是否有环
- 两个链表的第一个公共节点
- 复杂链表的复制
143. 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 may not modify the values in the list's nodes, only nodes itself may be changed.
Example 1:
Given 1->2->3->4, reorder it to 1->4->2->3.
Example 2:
Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
Hint: 链表中点 链表反转 链表合并
解题思路：
1.利用快慢指针找到链表中点
2.从中点处断开链表，分成 A、B 两段
3.反转中点之后的B段链表节点（注意细节，四步反转）
4.循环A、B 两段链表，并将第二段链表节点间隔插入第一段链表
/**
* 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 == nullptr || head->next == nullptr || head->next->next == nullptr) return;
ListNode* slow = head;
ListNode* fast = head;
while(fast->next && fast->next->next){
slow = slow->next;
fast = fast->next->next;
}
// 获得中点
ListNode* mid = slow->next;
// 从中点位置断开
slow->next = nullptr;
// 反转第二部分链表
ListNode* last = mid;
ListNode* prev = nullptr;
while(last){
ListNode* next = last->next;
last->next = prev;
prev = last;
last = next;
}
// 间隔插入第一个链表中
while(head && prev){
ListNode* next = head->next;
head->next = prev;
prev = prev->next;
head->next->next = next;
head = next;
}
}
};
