题解 | #反转链表#

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* ReverseList(ListNode* pHead) {
		if(pHead == nullptr){
			return nullptr;
		}

		ListNode* p = pHead;
		ListNode* q = p->next;
		p->next = nullptr;		// 头节点不为空时， 反转后，头节点的next应为空
		while (q != nullptr) {
			ListNode* k = q->next; 
			q->next = p;
			p = q;
			q = k;
		}
		return p;

    }
};