Given a linked list, remove the nth node from the end of list and return its head. Given n will always be valid and try to do this in one pass.

For example,

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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.

思路

这是一道非常典型的快慢指针题。双指针的操作在链表中比较常用，这样的操作可以省去频繁对链表进行遍历的开销，非常方便。

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# 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
        """
        fast = head
        headflag = 1
        # 利用快指针来定位倒数第n个元素
        for i in range(n):
            fast = fast.next
        # 慢指针必须要指到目标元素前一个元素
        # 才能执行单链表的删除操作
        if (fast):
            fast = fast.next
            headflag = 0
        slow = head
        while(fast):
            fast = fast.next
            slow = slow.next
        # 经过循环，现在慢指针已经指向目标元素的前一个元素了
        if (slow == head):
            # 边界条件：如果慢指针还在头部
            if (headflag):
                # 删除第一个元素
                head = slow.next
            else:
                slow.next = slow.next.next
        else:
            slow.next = slow.next.next
        return head