143. Reorder List

最新推荐文章于 2020-05-10 11:40:52 发布

原创最新推荐文章于 2020-05-10 11:40:52 发布 · 203 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

184 篇文章

订阅专栏

本文介绍了一种链表重构算法，该算法将链表分为两部分并反转后半部分，然后交错合并两部分链表。文章详细解释了如何寻找链表中点、反转链表以及交替合并两个链表的方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

class Solution {
public:
    void reorderList(ListNode* head) {
                // write your code here
        if(head == NULL || head->next == NULL){
            return;
        }

        ListNode *list1 = head;
        ListNode *mid = findmid(head);
        ListNode *list2 = reserve(mid->next);
        mid->next = NULL;
        ListNode *dummy = new ListNode(0);
        ListNode *temp = dummy;
        int index = 0;
        while(list1!=NULL && list2!=NULL){
            if(index%2 == 0){
                temp->next = list1;
                list1 = list1->next;
            }
            else{
                temp->next = list2;
                list2 = list2->next;
            }
            index++;
            temp = temp->next;
        }
        if(list1!=NULL){
            temp->next = list1;
        }
        if(list2!=NULL){
            temp->next = list2;
        }
        head = dummy->next;
    }
    ListNode *findmid(ListNode *head){
        ListNode *first = head;
        ListNode *second = head->next;
        while(second!=NULL && second->next!=NULL){
            first = first->next;
            second = second->next->next;
        }
        return first;
    }
    ListNode *reserve(ListNode *head){
        ListNode *prev = NULL;
        ListNode *cur = head;
        ListNode *temp;
        while(cur!=NULL){
            temp = cur->next;
            cur->next = prev;
            prev = cur;
            cur = temp;
        }
        return prev;
    }
 
};