92. Reverse Linked List II 反转链表 II

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<m-1; i++)
            temp = temp->next;
        pre = temp;
        first = pre->next;
        b = pre->next;
        c = b->next;
        cout << "b: " << b->val << endl;
        cout << "c: " << c->val << endl;
        for(; i<n-1; i++)
        {
            b->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;
    }
};