题目描述

法一）迭代法

class Solution{
public:
	ListNode* reverseList(ListNode* head) {
		ListNode* prev = NULL;
		ListNode* cur = head;
		while(cur){     
			ListNode* next = cur->next;
			cur->next = prev;
			prev = cur;
			cur = next; 
		}
		return prev;    //最后一步cur为空，prev为最后一个结点 
	}
};

法二）递归

class Solution{
public:
	ListNode* reverseList(ListNode* head) {
		if(!head || !head->next){   //结点为空或者只有一个结点
			return head; 	
		} 
		ListNode* newHead = reverseList(head->next);
		head->next->next = head;
		head->next=NULL;
		return newHead;
	}
};