Remove Nth Node From End

题目描述

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.

解题方法

遇到linked list的问题首先问自己

• head确定能保留吗？那么需要dummy node吗？

这一题head也不一定能够保留，所以要用dummy node来做，两个快慢指针的做法，快指针先跑出去n步

我们要删去 nth node from end, nth node到tail的距离是n-1, 而我们要获得是Nth node前面一个， 那么它跟tail的距离是n, 所以快指针应该先走出n步。

Solution

# 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 not head:
            return
        dummy = ListNode(0) # dummy node
        dummy.next = head
        p1 = dummy
        p2 = dummy
        for i in range(n):
            p2 = p2.next

        while p2 and p2.next:
            p1 = p1.next
            p2 = p2.next
        p1.next = p1.next.next

        return dummy.next

Reference