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.
分析
两个指针，一个先前进n，然后另一个一起移动，直到第一个达到尾部。
这题非要2个指针都从dummy起，从Head起就是不行，虽然做法完全一样。
因为可能删除的是头结点，所以一定要从dummy起。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode quick = dummy, slow = dummy;
for(int i = 0; i <= n; i++){
quick = quick.next;
}
while(quick != null){
slow = slow.next;
quick = quick.next;
}
slow.next = slow.next.next;
return dummy.next;
}
}