leetcode143 重排链表

思路：

先找到链表中点，再反转后半部分，再merge两个子链表。

实现：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution
{
public:
    ListNode* middleNode(ListNode* head)
    {
        if (head == NULL) return NULL;
        ListNode* fast = head, *slow = head;
        while (fast and fast->next)
        {
            fast = fast->next->next;
            slow = slow->next;
        }
        return slow;
    }
    ListNode* reverse(ListNode* head)
    {
        if (head == NULL or head->next == NULL) return head;
        ListNode* pre = head, *cur = head->next;
        while (cur)
        {
            ListNode* nxt = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nxt;
        }
        head->next = NULL;
        return pre;
    }
    void merge(ListNode* a, ListNode* b)
    {
        if (b == NULL) return;
        while (a)
        {
            ListNode* x = a->next;
            a->next = b;
            ListNode* y = b->next;
            if (x) b->next = x;
            a = x;
            b = y;
        }
    }
    void reorderList(ListNode* head)
    {
        if (head == NULL or head->next == NULL) return;
        ListNode* mid = middleNode(head);
        ListNode* tmp = head;
        while (tmp->next != mid) tmp = tmp->next;
        tmp->next = NULL;
        auto l = reverse(mid);
        merge(head, l);
    }
};