剑指 Offer 24. 反转链表

1. Question

定义一个函数,输入一个链表的头节点,反转该链表并输出反转后链表的头节点。

2. Examples

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

3. Constraints

  • 0 <= 节点个数 <= 5000

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

迭代,在原来的列表上改变next的指向。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode curr = head;

    while(curr != null) {
      ListNode next = curr.next;
      curr.next = prev;
      prev = curr;
      curr = next;
    }
    return prev;
  }
}
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