• 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;
            }
        }
    };
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  • 原文地址:https://www.cnblogs.com/ALINGMAOMAO/p/9886185.html
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