Leetcode解题206：反转链表

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题目：

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

分析：此题为链表的常规题，步骤概括为：备份head->next;修改head->next;移动head与new_head;

c++代码：

#include <stdio.h>

struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *new_head = NULL;
        while(head)
        {
            ListNode *next = head->next;
            head->next = new_head;
            new_head=head;
            head=next;
        }
        return new_head;
    }
};

int main()
{
	ListNode a(1);
	ListNode b(2);
	ListNode c(3);
	ListNode d(4);
	ListNode e(5);	
	a.next = &b;
	b.next = &c;
	c.next = &d;
	d.next = &e;
	e.next = NULL;
	
	Solution s;
	ListNode *head=&a;
	printf("反转前链表：\n");
	while(head)
	{
		printf("%d\n",head->val);
		head = head->next;
	}
	head=s.reverseList(&a);
	printf("反转后链表：\n");
	while(head)
	{
		printf("%d\n",head->val);
		head = head->next;
	}
}