[LeetCode] Reverse Linked List II

hale1007

于 2014-10-23 10:32:41 发布

阅读量329

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode Algorithm

本文链接：https://blog.youkuaiyun.com/hale1007/article/details/40392887

Algorithm 同时被 2 个专栏收录

86 篇文章

订阅专栏

LeetCode

76 篇文章

订阅专栏

本文介绍了一种算法，用于在单次遍历中就地反转链表的指定区间，从位置m到n。提供了完整的C++实现代码，并附带了链表节点的定义。通过实例演示了如何将链表的一部分进行反转。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 == nullptr || m == n)   return head;
        ListNode *p = head, *q = nullptr;
        for(int i = 0; i < m-1; i++)
        {
            q = p;
            p = p->next;
        }
        ListNode *end = p, *pre = p;
        p = p->next;
        for(int i = m+1; i <= n; i++)
        {
            ListNode *pNext = p->next;
            p->next = pre;
            pre = p;
            p = pNext;
        }
        end->next = p;
        if(q)
            q->next = pre;
        else
            head = pre;
        return head;
    }
};