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?
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链表翻转题
可以用O(1)的空间解决,就是新建立两个辅助结点,不占什么空间,然后,其中一个结点指向head,并且,head改为NULL,然后,进行翻转。
注意在循环是,这两个辅助结点的next的变化,不能漏掉,可以画图来辅助理解和检查。
emmm下面的代码不太好理解。用dummy->next指代head就好理解很多了。
C++代码:
不适用dummy.
/**
* 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) {
ListNode *p,*q;
p = head;
head = NULL;
while(p){
q = p->next;
p->next = head;
head = p;
p = q;
}
return head;
}
};
使用dummy.
/**
* 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) {
ListNode *dummy = new ListNode(-1);
dummy->next = head;
ListNode *p,*q;
p = dummy->next;
dummy->next = NULL;
while(p){
q = p->next;
p->next = dummy->next;
dummy->next = p;
p = q;
}
return dummy->next;
}
};
来源:https://www.cnblogs.com/Weixu-Liu/p/10703998.html