leetcode之-题19

# 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
        """
        """
        采用双指针思想，两个指针相隔n-1，每次两个指针向后一步，当后面一个指针没有后继了，前面一个指针就是要删除的节点
        """
        p = head
        q = head
        Ppre = None
        for x in xrange(0, n-1):
            q = q.next
        while q.next is not None:
            Ppre = p
            p = p.next
            q = q.next
        # 出现[1],n=1的情况下，Ppre依然为None
        if Ppre == None:
            head = p.next
        else:
            Ppre.next = p.next
        return head