reorder-list 单链表

最新推荐文章于 2025-09-10 15:24:29 发布

转载最新推荐文章于 2025-09-10 15:24:29 发布 · 65 阅读

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原文链接：http://www.cnblogs.com/ALINGMAOMAO/p/9886185.html

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#数据结构与算法

题意：

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