【每日一题】力扣之反转链表II

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        ListNode *sentinel = new ListNode(-1);
        sentinel->next = head;
        auto p = sentinel;
        for (int i = 0; i < left - 1; i++ )
        {
            p = p->next;
        }
        auto pre = p->next, post = p->next->next;
        for (int i = 0; i < right - left; i++ )
        {
            auto temp = post->next;
            post->next = pre;
            pre = post;
            post = temp;
        }
        p->next->next = post;
        p->next = pre;
        return sentinel->next;
    }
};