leetcode 19. 删除链表的倒数第N个节点 Remove Nth Node From End of List

使用双指针法，先将右指针右移N次。这时需要做一个判断，如果右指针已经到末尾，则说明待删除的是倒数第N个节点即第一个节点，直接返回head.next即可。否则，同时移动左指针和右指针，直到右指针达到末尾，此时左指针正好指向待删除节点的前一个位置。

python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        left = head
        right = head
        for i in range(n):
            right = right.next
        if(right is None):  #说明要删除的是第一个节点
            return head.next
        while(right.next):
            left = left.next
            right = right.next
        left.next = left.next.next
        return head