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

核心部分：通过a、b、c三个变量之间的相互更新，不断反转部分链表

然后将反转部分左右两端接上！

当测试数据 m 为 1 时，原始代码行不通。

故我们在原head前加一个fake_h节点，在函数部分将m++，n++，最后return fake_h->next

注意判断head为空 和 不反转任何部分（m==n）这两种情况

/** * 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->next || m==n)        {            return head;        }        m++;        n++;        ListNode* pre = NULL;        ListNode* first = new ListNode(0);        ListNode* a = NULL;        ListNode* b = new ListNode(0);        ListNode* c = new ListNode(0);        ListNode* fake_h = new ListNode(0);        fake_h->next = head;        ListNode* temp = fake_h;        int i = 1;        for(; i
     
      next;        pre = temp;        first = pre->next;        b = pre->next;        c = b->next;        cout << "b: " << b->val << endl;        cout << "c: " << c->val << endl;        for(; i
      
       next = a;            a = b;            cout << "a: " << a->val << endl;            b = c;            c = c->next;        }        b->next = a;        cout << "pre: " << pre->val << endl;        pre->next = b;        first->next = c;        return fake_h->next;    }};