原題地址:http://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/
題意:
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.
這道題的含義是刪除鏈表的倒數第n個節點。
解題思路:加一個頭結點dummy,並使用雙指針p1和p2。p1先向前移動n個節點,然后p1和p2同時移動,當p1.next==None時,此時p2.next指的就是需要刪除的節點。
代碼:
# 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