4.反转指定位置段的链表

题目描述

给你单链表的头指针head和两个整数left和right,其中 left<=right。请你反转从位置left到位置right的链表节点，返回反转后的链表。

题目链接

https://leetcode-cn.com/problems/reverse-linked-list-ii/description/

样例

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]
 
输入：head = [5], left = 1, right = 1
输出：[5]

进阶

使用一趟扫描完成

时间复杂度分析

整个链表只遍历了一遍，所以时间复杂度为O(n)

理解

1.需要考虑的是，如果链表中只有两个结点的情况。
2.定义一个虚拟结点，指向头结点

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        ListNode* fictitious = new ListNode(0);
        fictitious->next = head;
 
        if (left == right) {
            return head;
        }
 
        int num = left-1;
        ListNode* first = fictitious;
        ListNode* last = first->next;
        while (num--) {
            first = first->next;
            last = first->next;
        }
        int sum = right - left;
        ListNode* a = last;
        ListNode* b = a->next;
        
        while (sum--) {
            ListNode* c = b->next;
            b->next = a;
            a = b;
            b = c;
        }
        first->next = a;
        last->next = b;        
        return fictitious->next;
    }
};