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 || !(pHead->next)) return pHead;
        //申请两个指针，一个记录前驱结点，一个记录后继结点
        ListNode* q = NULL;
        ListNode* r = pHead->next;
        while (pHead) {
            pHead->next = q;
            q = pHead;
            pHead = r;
            //r为空时，再后移，会报错，
            if (r) r = r->next;
        }
        return q;
    }
};