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?
class Solution {
public:
ListNode* reverseList(ListNode* head) {
//递归
ListNode* newHead = NULL;
while (head) {
ListNode* nextNode = head->next;
head->next = newHead;
newHead = head;
head = nextNode;
}
return newHead;
}
};
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseListIter(ListNode* head, ListNode* newHead) {
//迭代
if (head == NULL) return newHead;
ListNode* nextNode = head->next;
head->next = newHead;
return reverseListIter(nextNode, head);
}
ListNode* reverseList(ListNode* head) {
return reverseListIter(head, NULL);
}
};