[leetcode]19. Remove Nth Node From End of List

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.

主要方法就是设置两个指针，第一个指针领先第二个n步，当第一个指针到头时去掉第二个指针所在的节点。在给的链表之前加一个头结点，方便遍历。

java代码

/** * 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 pre = new ListNode(0);        pre.next = head;        ListNode ahead = pre,behind = pre;        for(int i=0;i<n;i++){            ahead=ahead.next;        }        while(ahead.next!=null){            ahead=ahead.next;            behind=behind.next;        }        behind.next=behind.next.next;        return pre.next;            }}

golang代码

/** * Definition for singly-linked list. * type ListNode struct { *     Val int *     Next *ListNode * } */func removeNthFromEnd(head *ListNode, n int) *ListNode {    pre:=&ListNode{        Val:0,        Next:head,    }    ahead:=pre    behind:=pre    for i:=0;i<n;i++{        ahead=ahead.Next    }    for ahead.Next!=nil{        ahead=ahead.Next        behind=behind.Next    }    behind.Next=behind.Next.Next    return pre.Next    }