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 = None

class 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;
}
}