[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}.

OH! MY GOD! I HATE LINKED LIST!

看似简单，实现起来总会遇到各种指针错误，写程序之前最好先在纸上好好画画，把各种指针关系搞搞清楚。

本题的想法就是先将列表平分成两份，后一份逆序，然后再将两段拼接在一起，逆序可用头插法实现。注意这里的函数参数要用引用，否则无法修改指针本身的值！

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void reverseList(ListNode *&head) {
        ListNode *h = new ListNode(0);
        ListNode *tmp;
        while (head != NULL) {
            tmp = head->next;
            head->next = h->next;
            h->next = head;
            head = tmp;
        }
        head = h->next;
    }

    void twistList(ListNode *&l1, ListNode *&l2) {
        ListNode *p1, *p2, *tmp;
        p1 = l1; p2 = l2;
        while (p1 != NULL && p2 != NULL) {
            tmp = p2->next;
            p2->next = p1->next;
            p1->next = p2;
            p1 = p1->next->next;
            p2 = tmp;
        }
    }   

    void reorderList(ListNode *head) {
        if (head == NULL || head->next == NULL || head->next->next == NULL) {
            return;
        }
        ListNode *slow, *fast;
        slow = head; fast = head;
        while (fast != NULL && fast->next != NULL) {
            slow = slow->next;
            if (fast->next->next == NULL) {
                break;
            }
            fast = fast->next->next;
        }
        ListNode *l2 = slow->next;
        slow->next = NULL;
        reverseList(l2);
        twistList(head, l2);
    }
};