算法题--重排链表

难度系数 中等 通过率 24%
描述
笔记
数据
评测

给定一个单链表L: L0→L1→…→Ln-1→Ln,

重新排列后为:L0→Ln→L1→Ln-1→L2→Ln-2→…

必须在不改变节点值的情况下进行原地操作。

您在真实的面试中是否遇到过这个题? Yes 样例 给出链表 1->2->3->4->null,重新排列后为1->4->2->3->null


/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
 
 
class Solution {
public:
    /*
     * @param head: The head of linked list.
     * @return: nothing
     */
    void reorderList(ListNode * head) {
        // write your code here
            if (head == nullptr || head->next == nullptr)
            {
                return;
            }
         
            ListNode *slow = head, *fast = head->next;
         
            while(fast&&fast->next)
            {
                slow = slow->next;
                fast = fast->next->next;
            }
         
            fast = slow->next;
            slow->next = nullptr;
            ListNode *rHead = nullptr;
            while (fast)
            {
                ListNode *r = fast->next;
                fast->next = rHead;
                rHead = fast;
                fast = r;
            }
         
         
            fast = rHead;
            slow = head;
            while(slow&&fast)
            {
                ListNode *rr = fast->next;
                ListNode *lr = slow->next;
                fast->next = lr;
                slow->next = fast;
         
                fast = rr;
                slow = lr;
            }
    }
};
    原文作者:LTELTY
    原文地址: https://blog.csdn.net/qq_25026989/article/details/87724004
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞