[leetcode]19. Remove Nth Node From End of List

Posted 2021-12-09 alfredsun

链表题没法在本地打断点调试真的烦。。。不过好在这题简单

注意考虑要去除的元素在正数第一位的情况就好。

 

Runtime: 40 ms, faster than 88.15% of Python3 online submissions forRemove Nth Node From End of List.

Memory Usage: 13.1 MB, less than 64.23% of Python3 online submissions for Remove Nth Node From End of List.

 

Submission Detail

208 / 208 test cases passed.	Status:  Accepted
Runtime: 40 ms Memory Usage: 13.1 MB	Submitted: 1 minute ago

 

# 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:
        listval =[]
        temp = head
        ret = head
        while(temp.next != None):
            listval.append(temp.val)
            temp = temp.next
        #lastone
        listval.append(head.val)
        lengthS = len(listval)
        #0
        if lengthS == 0:
            return head
        #1
        if lengthS == 1:
            return []
        #normal
        reverse_n = lengthS+1 -n
        #reverse_n == 1
        if reverse_n ==1:
            return head.next
        for index in range(1,reverse_n):
            if index+1 == reverse_n:
                head.next = (head.next).next
            else:
                head = head.next
        return ret

 

 

16ms：

# 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:
        cur = head
        for i in range(n):
            cur = cur.next
        if not cur:
            head = head.next
            return head
        else:
            pre = head
            cur = pre.next
        
        while cur:
            for i in range(n):
                cur = cur.next
            if not cur:
                pre.next = pre.next.next
                return head
            else:
                pre = pre.next
                cur = pre.next