[LeetCode 206] 反转链表 I

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💭 解题思路#

递归#

方法的意思是反转链表，我们直接反转head.next。然后再把head做一下反转即可。

class Solution {
    public ListNode reverseList(ListNode head) {
      //base case 如果节点为空节点，或者单节点，直接返回即可，不用反转
      if (head == null || head.next == null)return head;

      //反转
      ListNode temp = reverseList(head.next);
      //反转head
      head.next.next = head;
      head.next = null;
      return temp;
    }
}

迭代#

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