143. 重排链表

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1:

给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例 2:

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reorder-list
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快慢指针找到链表中点，将前后部分断开，后半部分反转，然后依次插入前半部分

//快慢指针找链表中点，slow就是链表中点
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast->next && fast->next->next){
            slow = slow->next;
            fast = fast->next->next;
        }

    ListNode* reverseList(ListNode* head) {
        ListNode* ret = nullptr;
        while (head){
            ListNode* next = head->next;
            head->next = ret;
            ret = head;
            head = next;
        }
        return ret;
    }

/**
 * 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:
    void reorderList(ListNode* head) {
        if (!head || !head->next) {
            return;
        }

        //快慢指针找链表中点，slow就是链表中点
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast->next && fast->next->next){
            slow = slow->next;
            fast = fast->next->next;
        }

        //后半部分头节点
        ListNode* later = slow->next; 
        //断开前后部分
        slow->next = nullptr;
        //反转后半部分
        later = reverseList(later);


        ListNode* front = head;
        while (front && later) {
            ListNode* frontNext = front->next;
            ListNode* laterNext = later->next;
            front->next = later;
            later->next = frontNext;
            front = frontNext;
            later = laterNext;
        }
    }

    //反转链表
    ListNode* reverseList(ListNode* head) {
        ListNode* ret = nullptr;
        while (head){
            ListNode* next = head->next;
            head->next = ret;
            ret = head;
            head = next;
        }
        return ret;
    }
};