剑指 Offer 24. 反转链表

迭代

class Solution {public ListNode reverseList(ListNode head) {ListNode pre = null, cur = head;while(cur != null){ListNode next = cur.next;cur.next = pre;pre = cur;cur = next;}return pre;}
}

递归（使用额外参数）

和上面迭代的思想一样。

class Solution {public ListNode reverseList(ListNode head) {return reverse(head, null);}ListNode reverse(ListNode cur, ListNode pre){if(cur == null) return pre;ListNode res = reverse(cur.next, cur);cur.next = pre;return res;}
}

递归（不使用额外参数）

class Solution {public ListNode reverseList(ListNode head) {if(head == null || head.next == null) return head; ListNode last = reverseList(head.next);head.next.next = head;head.next = null;return last;}
}