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;
}
}