19. Remove Nth Node From End of List

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.

AC代码：

#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):
        front = behind = head
        for _ in xrange(n):
            front = front.next
        if front == None: return behind.next
        while front.next:
            front = front.next
            behind = behind.next
        behind.next = behind.next.next
        return head

思路：

因为是移除从后面数第n个，所以用两个指针，第一个先移动n次，然后两个指针一起移动，快的指针移动到最后的时候，慢的指针正好移动到离队尾差n个。