leetcode第19题——*Remove Nth Node From End of List

题目

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.

思路

先得出链表长度len，然后根据n和len找出要删除的链表节点，比如要删除的节点是curr，前面的节点是prev，则prev.next=curr.next则将该节点删除。

Python语言也可以遍历链表将当前节点curr的上一节点prev存入一个list里面，按照以下思路便可以不用求得链表长度：如果n=1那么list只存储一个节点（即prev）；如果n=2那么list存储两个节点（即prev和prev的上一节点）...以此类推，最后list里的第一个节点便是要删除节点的前节点.

代码

Python

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def removeNthFromEnd(self, head, n):        """        :type head: ListNode        :type n: int        :rtype: ListNode        """        self.next, nodelist = head, [self]        while head.next != None:            if len(nodelist) == n:                nodelist.pop(0)            nodelist += head,            head = head.next        nodelist[0].next = nodelist[0].next.next        return self.next

Java

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        ListNode temp = head;int cnt = 0;while (temp != null) {//计算链表长度cnt++;temp = temp.next;}if (cnt < 2) return null;ListNode prev = null;cnt = cnt - n;temp = head;while (cnt > 0 && temp != null) {//找到要删除的节点prev = temp;temp = temp.next;cnt--;}if (temp != null) {if (prev != null) prev.next = temp.next;else return head.next;//要删除的是头节点}return head;    }}