Leetcode Reverse Linked List II

最新推荐文章于 2025-12-21 17:04:13 发布

原创最新推荐文章于 2025-12-21 17:04:13 发布 · 190 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

124 篇文章

订阅专栏

本文介绍了一种在链表中从位置m到n进行逆序的方法，并提供了一个C++实现示例。该方法通过一次遍历完成逆序操作，适用于1≤m≤n≤链表长度的情况。

Reverse a linked list from position m ton. 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.

看到题目的第一想法，就是经典的链表逆序的变形，思想简单。

代码如下：

/**
 * 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(n==m)
            return head;
        ListNode* curr = new ListNode(0);
        ListNode* pre = curr;
        curr->next = head;
        int count = 1;
        while(count < m)
        {
            curr = curr->next;
            count++;
        }
        ListNode* thead = curr;
        ListNode* temp;
        curr = curr->next->next;
        count++;
        while(count <= n)
        {
            temp = curr->next;
            curr->next = thead->next;
            thead->next = curr;
            curr = temp;
            count++;
        }
        count=0;
        while(count < n-m+1)
        {
            thead = thead->next;
            count++;
        }
        if(thead != NULL)
            thead->next = curr;
        return pre->next;
    }
};

代码有点混乱，进了整理后如下：

/**
 * 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(n==m)
            return head;
        n -= m;
        ListNode* thead = new ListNode(0);
        ListNode* pre = thead;
        thead->next = head;
        while(--m)
            thead = thead->next;
  
        ListNode* temp;
        ListNode* pstart = thead->next;

        while(n--)
        {
            temp = pstart->next;
            pstart->next = temp->next;
            temp -> next = thead->next;
            thead->next = temp;
        }

        return pre->next;
    }
};