[数据结构与算法]特殊翻转链表的相同解法

//k个链表翻转
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode *dummy = new ListNode(-1), *pre = dummy, *cur = pre;
        dummy->next = head;
        int num = 0;
        while(cur->next){
            num++;
            cur = cur->next;
        }
        while(num >= k){
            cur = pre->next;
            for(int i = 1; i < k; i++){
                ListNode *t = cur->next;
                cur->next = t->next;
                t->next = pre->next;
                pre->next = t;
            }
            pre = cur;
            num -= k;
        }    
        return dummy->next;
        
    }
};

//翻转n到m的链表
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode *dummy = new ListNode(-1), *pre = dummy;
        dummy->next = head;
        for (int i = 0; i < m - 1; ++i) 
            pre = pre->next;
        
        ListNode *cur = pre->next;
        for(int i = m; i < n; ++i){
            ListNode *t = cur->next;
            cur->next = t->next;
            t->next = pre->next;
            pre->next = t;
        }
        
        return dummy->next;
    }
};