Leetcode92.Reverse_Linked_List_II

以1->2->3->4->5(m=2,n=4)为例



时间复杂度:O(N)
C++代码：

class Solution {
public:
	ListNode * reverseBetween(ListNode* head, int m, int n) {
		if (head == nullptr || head->next == nullptr || m == n)
			return head;
		ListNode* dummy = new ListNode(-1);
		dummy->next = head;
		ListNode* frontMark = dummy, *rearMark, *p1, *p2, *p3;
		for (int i = 0; i < m - 1; i++)
			frontMark = frontMark->next;
		p1 = frontMark->next;
		rearMark = p1;
		p2 = p1->next;
		p3 = p2->next;
		int count = 0;
		while (count < n - m)
		{
			p2->next = p1;
			p1 = p2;
			p2 = p3;
			if (p3 == nullptr)
				break;
			p3 = p3->next;
			count++;
		}
		frontMark->next = p1;
		rearMark->next = p2;
		return dummy->next;
	}
};