刷题206. Reverse Linked List

一、题目说明

题目206. Reverse Linked List，翻转一个链表。难度是Easy！可以递归，也可以非递归。

二、我的解答

翻转，这个确实不难，非递归算法：

class Solution{
	public:
		ListNode* reverseList(ListNode* head){
			if(head == NULL || head->next==NULL) return head;
			ListNode *cur,*p = head->next;
			bool first = true;
			while(p !=NULL && p->next!=NULL){
				cur = p->next;
				
				p->next = head;
				if(first){
					head->next = NULL;
					first = false;
				}
				
				head = p;
				p = cur;
			}
			if(p !=NULL){
				p->next = head;
				if(first){
					head->next = NULL;
					first = false;
				}
				head = p;
			}
			
			return head;
		}
};

性能如下：

Runtime: 8 ms, faster than 88.11% of C++ online submissions for Reverse Linked List.
Memory Usage: 9.7 MB, less than 5.34% of C++ online submissions for Reverse Linked List.

三、优化措施

用翻转链表法实现：

class Solution{
	public:
		ListNode* reverseList(ListNode* head){
			//翻转链表 
			ListNode *pre = NULL;
			ListNode *cur = head;
			ListNode *tmp = NULL;
			while(cur!=NULL) {
				//记录当前节点的下一个节点
				tmp = cur->next;
				//然后将当前节点指向pre
				cur->next = pre;
				//pre和cur节点都前进一位
				pre = cur;
				cur = tmp;
			}
			return pre;
		}
};

Runtime: 12 ms, faster than 38.84% of C++ online submissions for Reverse Linked List.
Memory Usage: 9.7 MB, less than 5.34% of C++ online submissions for Reverse Linked List.