Leet Code 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.

【算法思路】

用两个指针，相差距离为n，先移动一个指针n次，然后再同时移动，当后一个指针为尾指针时，前一个指针指向了要删除元素的前一个节点。

【复杂度】

时间：O（n）

public ListNode removeNthFromEnd(ListNode head, int n) {ListNode deleteNode = head;ListNode tailNode = head;if((head.next == null) && (n >= 1)){return null;}for(int i = 0 ; i < n; i ++){if(tailNode.next != null)tailNode = tailNode.next;else {if(i == n - 1)head = head.next;return head;}}while(tailNode.next != null){deleteNode = deleteNode.next;tailNode = tailNode.next;}tailNode = deleteNode.next;deleteNode.next = tailNode.next;tailNode.next = null;tailNode = null;return head;}