[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

}