Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:
Could you do this in one pass?

难度:medium

题目:
给定一链表,移除其倒数第n个结点。
注意:n总是合法数

思路:双指针

Runtime: 6 ms, faster than 98.72% of Java online submissions for Remove Nth Node From End of List.
Memory Usage: 27 MB, less than 39.84% of Java online submissions for Remove Nth Node From End of List.

/**
 * 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 dummyHead = new ListNode(0);
        dummyHead.next = head;
        ListNode ptr = dummyHead,lastNPtr = dummyHead;
        while (ptr.next != null) {
            if (--n < 0) {
                lastNPtr = lastNPtr.next;
            }
            
            ptr = ptr.next;
        }
        
        lastNPtr.next = lastNPtr.next.next;
        
        return dummyHead.next;
    }
}

linm
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