算法1

把那一段放进数组里，后一个指向前一个，边界值处理一下就好
搞一个空的头指针方便点

时间复杂度 $O(n)$

C++ 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 *tmp = new ListNode(0,head);
        int index = 0;
        ListNode *p = tmp;
        while(index < left -1)
        {
            p = p->next;
            index++;
        }
        vector<ListNode *> vec;
        while(index<=right)
        {
            vec.push_back(p);
            p = p->next;
            index++;
        }
        ListNode *n = vec[vec.size()-1]->next;
        for(int i=2;i<vec.size();i++)
        {
            vec[i]->next = vec[i-1];
        }
        vec[1]->next = n;
        vec[0]->next = vec[vec.size()-1];
        return tmp->next;
    }
};