Reverse Linked List（LeetCode）

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Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

反转链表，有两种方式：迭代和递归

方法1：迭代

思路：用三个辅助指针，从头结点开始遍历反向，最后加上空结点

struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};
//Iterative
class Solution {
public:
	ListNode* reverseList(ListNode* head) {
		if (head == NULL || head->next == NULL)
			return head;
		ListNode *p1, *p2, *p3;
		p1 = head, p2 = head->next;
		while (p2)
		{
			p3 = p2->next;
			p2->next = p1;//转向
			p1 = p2;//前进
			p2 = p3;
		}
		head->next = NULL;
		head = p1;
		return head;
	}
};

方法2：递归

思路：找到最后一个结点，反向，后面有序了就断链不用排了，进入下一次递归。

附上一位印度小哥的图解：https://www.youtube.com/watch?v=KYH83T4q6Vs

class Solution {
public:
	ListNode* reverseList(ListNode* head) {
		if (head == NULL || head->next == NULL)
			return head;
		ListNode *p = head;
		head = reverseList(p->next);
		p->next->next = p;//反向
		p->next = NULL;//unlink
		return head;
	}
};