leetcode206 链表的反转

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?

思路：每一次翻转 当前节点的下一个结点指向它的前一个结点；它的下一个结点的指针指向它；

ListNode *cur=head;

ListNode *pre=nullptr;

ListNode *pNext=cur->next;

反转的时候：cur->next=pre;pre=cur;cur=pNext(原来的cur->next);

然后 cur反复的循环 从头往后遍历；

CODE：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) 
    {
        if(head==nullptr)
            return nullptr;
        ListNode *pre=nullptr;
        ListNode *cur=head;
        while(cur)
        {
            ListNode *pNext=cur->next;
            cur->next=pre;
            pre=cur;
            cur=pNext;
        }
        return pre;
    }
};