reorder-list 单链表

题意:

  给你一个单链表 a1 a2 a3 a4 a5....an  让你变成 a1 an a2 an-1 a3 an-2 ....

这里牵涉到,单链表的倒置和两个单链表的合并。

class Solution {
public:
    void reorderList(ListNode *head) {
        if (!head || !head->next)return;    //空表和只有一个元素的链表不用操作
        //快慢指针找中间位置
        ListNode *fast = head;        ListNode *slow = head;
        while (fast&&fast->next)
        {
            slow = slow->next;    fast = fast->next->next;
        }
        //将后半段的链表倒置
        ListNode *after = slow->next;    slow->next = NULL;
        ListNode *pre = NULL;
        while (after){
            ListNode *p = after->next;
            after->next = pre;
            pre = after;    after = p;
        }
        //合并两个链表
        ListNode *first = head;
        after = pre;
        while (first&&after){
            ListNode *ftemp = first->next;
            ListNode *aftemp = after->next;
            first->next = after;    first = ftemp;
            after->next = first;    after = aftemp;
        }
    }
};
原文地址:https://www.cnblogs.com/ALINGMAOMAO/p/9886185.html