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
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题?来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reverse-linked-list
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解法:
/**
* 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 == NULL) return NULL;
ListNode *curr = NULL;
while(head)
{
ListNode *res = head->next;
head->next = curr;
curr = head;
head = res;
}
return curr;
}
};
解法二:
class Solution {
public:
ListNode* reverseList(ListNode* head)
{
if (!head || !head->next) return head;
ListNode *newHead = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return newHead;
}
};