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 pHead;

		ListNode * left=nullptr;
		ListNode * cur = pHead;
		ListNode * right =pHead->next;
		//while right ==null,stop
		//if not,process cur ->left,
		//move left, cur,right,
		while(right!=nullptr){
			cur->next=left;

			left=cur;
			cur=right;
			right=right->next;
		}

		cur->next=left; //the last node

		return cur;
    }
};

1.便利结束后，还剩末端没弯转

2.过早翻转会丢失后续的node，因此应该提前记下下一个node