题目描述
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？

/*** 思路1：迭代法，直接依次反转链表* 时间复杂度O(n),空间复杂度O(1)* @param head* @return*/public ListNode reverseList(ListNode head) {if(head == null || head.next == null) {return head;}ListNode curr = head, next = head.next, nNext = next;curr.next = null;while(nNext != null) {nNext = nNext.next;next.next = curr;if(nNext != null) {curr = next;next = nNext;}}return next;}

思路2：递归法

/*** 思路2：递归解决* 时间复杂度O(n),空间复杂度O(1)* @param head* @return*/public ListNode reverseList(ListNode head) {if(head == null || head.next == null) {return head;}ListNode curr = head, next = head.next, nNext = next.next;curr.next = null;return one(curr, next, nNext);}private ListNode one(ListNode curr, ListNode next, ListNode nNext) {next.next = curr;if(nNext != null) {return one(next, nNext, nNext.next);}else {return next;}}