[leetcode]Remove Nth Node From End of List @ Python
2014-04-30 18:42
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原题地址:http://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/
题意:
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Note:
Given n will always be valid.
Try to do this in one pass.
这道题的含义是删除链表的倒数第n个节点。
解题思路:加一个头结点dummy,并使用双指针p1和p2。p1先向前移动n个节点,然后p1和p2同时移动,当p1.next==None时,此时p2.next指的就是需要删除的节点。
代码:
题意:
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.
这道题的含义是删除链表的倒数第n个节点。
解题思路:加一个头结点dummy,并使用双指针p1和p2。p1先向前移动n个节点,然后p1和p2同时移动,当p1.next==None时,此时p2.next指的就是需要删除的节点。
代码:
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @return a ListNode def removeNthFromEnd(self, head, n): dummy=ListNode(0); dummy.next=head p1=p2=dummy for i in range(n): p1=p1.next while p1.next: p1=p1.next; p2=p2.next p2.next=p2.next.next return dummy.next
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