19. Remove Nth Node From End of List

题目：

Given a linked list, remove the n^th 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.

代码：

从链表中删除从右向左数第N个元素，况且只能遍历链表一遍。

所以，需要定义两个指针，相差是N个元素，同时向前遍历，当靠前的元素走到最后的时候，在后面的元素刚好与最后一个元素距离为N，删除即可。

于是：

#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

        """

        p = head

        q = head

        if head.next ==None:return []

        for i in range(0,n):

            q = q.next

        if not q : return head.next

        while q.next:

            p = p.next

            q = q.next

        temp = p.next.next

        p.next = temp

        return head

if __name__=='__main__':

    arr = [1,2]

    p = ListNode(arr[0])

    head = p

    for i in arr[1:]:

        p.next = ListNode(i)

        p = p.next

    s=Solution()

    q=s.removeNthFromEnd(head,2)

    while q:

        print (q.val)

        q = q.next

巴特西

19. Remove Nth Node From End of List

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