方法一:迭代法
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode*pre=NULL; //当前节点的前驱节点
ListNode*now=head; //当前节点
while(now!=NULL){
ListNode*next=now->next; //记住当前节点的后继节点,防止丢失后面的链表
//接下来处理当前节点,反转指针指向
now->next=pre; //让当前节点的next指针指向pre
pre=now; //pre指针后移
now=next; //now指针后移
}
return pre;
}
};
标签:pre,ListNode,206,next,链表,节点,now,leetcode
From: https://www.cnblogs.com/cxyupup/p/17331322.html