leetcode——19. 删除链表的倒数第N个节点

用列表完成！！！！

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        if head.next==None and n==1:
            return 
        a=[]
        while head:
            a.append(head.val)
            head=head.next
        #print(a)
        #print(len(a))
        a.pop(len(a)-n)
        head=ListNode(a[0])
        p=head
        for i in range(1,len(a)):
            p.next=ListNode(a[i])
            p=p.next
        return head

执行用时 :44 ms, 在所有 python3 提交中击败了86.21%的用户

内存消耗 :13.8 MB, 在所有 python3 提交中击败了5.53%的用户

 

                                                                              ——2019.10.23

 

---

public ListNode removeNthFromEnd(ListNode head, int n) {
        if(n == 0){
            return head;
        }
        if(n == 1 && head.next == null){
            return null;
        }
        if(n == 2 && head.next.next == null){
            return head.next;
        }
        ListNode dummy = new ListNode(-1);
        ListNode p = dummy;
        dummy.next = head;
        ListNode t = p;
        ListNode prev = p;
        while (n != 0){
            p = p.next;
            n--;
        }
        while (p != null){
            p = p.next;
            t = t.next;
            if(t != head){
                prev = prev.next;
            }
        }
        prev.next = t.next;
        t.next = null;
        return dummy.next;
    }

——2020.7.14