LeetCode 1474. 删除链表 M 个节点之后的 N 个节点

输入: head = [1,2,3,4,5,6,7,8,9,10,11,12,13], m = 2, n = 3
输出: [1,2,6,7,11,12]
解析: 保留前(m = 2)个结点,  也就是以黑色节点表示的从链表头结点开始的结点(1 ->2).
删除接下来的(n = 3)个结点(3 -> 4 -> 5), 在图中以红色结点表示.
继续相同的操作, 直到链表的末尾.
返回删除结点之后的链表的头结点.

输入: head = [1,2,3,4,5,6,7,8,9,10,11], m = 1, n = 3
输出: [1,5,9]
解析: 返回删除结点之后的链表的头结点.

示例 3:
输入: head = [1,2,3,4,5,6,7,8,9,10,11], m = 3, n = 1
输出: [1,2,3,5,6,7,9,10,11]

示例 4:
输入: head = [9,3,7,7,9,10,8,2], m = 1, n = 2
输出: [9,7,8]
 
提示:
 1 <= 链表结点数 <= 10^4.
[1 <= 链表的每一个结点值 <=10^6].
1 <= m,n <= 1000

class Solution {
public:
    ListNode* deleteNodes(ListNode* head, int m, int n) {
        int count = m;
        ListNode* cur = head, *tail = NULL;
        while(cur)
        {
        	count = m;
        	while(cur && count--)
        	{
        		tail = cur;
        		cur = cur->next;
        	}
        	count = n;
        	while(cur && count--)
        	{
        		cur = cur->next;
        	}
        	tail->next = cur;
        }
        tail->next = NULL;
        return head;
    }
};