92. 反轉鏈表 II

就用1，2，3，4，5 畫個圖就行。

 class Solution {
 public:
	 ListNode* reverseBetween(ListNode* head, int m, int n) {
		 if (m > n || m < 0)return nullptr;
		 ListNode *dummy = new ListNode(0);
		 dummy->next = head;
		 ListNode *p = dummy;
		 int count = 1;
		 while (count<m)
		 {
			 count++;
			 p = p->next;
		 }
		 reverse(p, n - m + 1);//以p開始翻轉n-m+1
		 return dummy->next;
	 }
	 ListNode *reverse(ListNode *preStart, int step)
	 {
		 ListNode *last = preStart->next;//preStart是第m個點的前一個，此時last指向第m個點
		 if (last == nullptr)return preStart;//這裏要注意第m個點不存在
		 ListNode * M_head = last;
		 ListNode * cur = last->next;
		 ListNode *tmp = nullptr;
		 while (step>1 && cur != nullptr)
		 {
			 tmp = cur->next;
			 cur->next = last;

			 last = cur;
			 cur = tmp;
			 step--;
		 }
		 M_head->next = cur;//第m個點
		 preStart->next = last;
		 return M_head;
	 }
 };