[leetcode] 19. Remove Nth Node From End of List python實現【easy】
阿新 • • 發佈:2019-01-25
- Remove Nth Node From End of List My Submissions QuestionEditorial Solution
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
意思是說給你一個單鏈表, 然後讓你刪掉從後往前的第n個,比如這個
而且有要求,最好跑一次。。
那想法就是記錄一下唄,每次跑的時候都記一下它前第n個位置那個節點。如果當前節點變成了空(也就是到了結尾),則刪掉對應的記錄的那個節點(前第n個節點)。
雙指標的意思
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
if head == None:
return
p = q = head
r = None
i = 0
while q != None:
if i != n:
i +=1
q = q.next
continue
else:
r = p
p = p.next
q = q.next
if r ==None:
return head.next
r.next = p.next
return head