LeetCode || Reverse Linked List II

最新推荐文章于 2022-10-28 13:04:35 发布

原创最新推荐文章于 2022-10-28 13:04:35 发布 · 332 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Reverse Linked List #Linked List #LeetCode

LeetCode 专栏收录该内容

107 篇文章

订阅专栏

本文介绍了如何在不使用额外空间的情况下，通过一次遍历反转链表中的指定区间。具体而言，从给定的链表头节点开始，通过迭代找到反转区间的起始和结束节点，并实现区间内的链表反转，最终返回更新后的链表头。

Reverse Linked List II

题目链接

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if(head == NULL || head->next == NULL ||  m > n || m < 1 || n < 1)
            return head;
        
        ListNode* insertp = NULL;//表示所需要插入节点的位置
        int count = 1;//计数器，从m到n
        ListNode* curr = head;
        while(curr != NULL && count < m)
        {
            insertp = curr;
            curr = curr->next;
            count++;
        }
        
        if(curr == NULL)
            return head;
        
        //curr记住了m所在的位置，接下来从m-n之间将所有nodes reverse
        ListNode* pre = curr;
        curr = curr->next;
        count ++;
        while(curr != NULL && count <= n)
        {
            if(insertp == NULL)//反复往头结点插入
            {
                ListNode* tmp2 = curr->next;
                
                ListNode* tmp = head;
                head = curr;
                head->next = tmp;
                
                pre->next = tmp2;
                curr = tmp2;
                count++;
                continue;
            }
            //put curr to [insertp, insertp->next]
            ListNode* tmp2 = curr->next;
            ListNode* tmp = insertp->next;
            insertp->next = curr;
            curr->next = tmp;
            //delete curr
            pre->next = tmp2;
            
            curr = tmp2;
            count++;
        }
        
        return head;
        
    }
};