Reorder List

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}.

中文：将链表重新排序，一个是升序，一个是降序。

分析：
思路就是：

1 用快慢指针找到中间节点
2 翻转中间节点后一个元素到最后一个元素区间的所有元素
3 断开前半段和翻转后的后半段元素
4 把前半段和翻转后的后半段元素以交叉的方式合并起来
5 特殊处理输入为空，只有一个元素和只有两个元素的corner case，就是多加几个if…return

/**
 * 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) return;
ListNode *slow = head, *fast = head, *prev = nullptr;
while (fast && fast->next) {
prev = slow;
slow = slow->next;
fast = fast->next->next;
}
prev->next = nullptr; // cut at middle
slow = reverse(slow);
// merge two lists
ListNode *curr = head;
while (curr->next) {
ListNode *tmp = curr->next;
curr->next = slow;
slow = slow->next;
curr->next->next = tmp;
curr = tmp;
}
curr->next = slow;
}
ListNode* reverse(ListNode *head) {
if (head == nullptr || head->next == nullptr) return head;
ListNode *prev = head;
for (ListNode *curr = head->next, *next = curr->next; curr;
prev = curr, curr = next, next = next ? next->next : nullptr) {
curr->next = prev;
}
head->next = nullptr;
return prev;
}
};