LeetCode – Remove Nth Node From End of List (Java)

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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.

注意题目要求一趟完成（也就是只遍历一次），自己写了个两趟完成的方法如下：

public class ListNode {int val;ListNode next;ListNode(int x) {val = x;}}

public class RemoveNodeFromList {//两趟完成，遍历两次public ListNode removeNthFromEnd(ListNode head, int n) {if(head == null) {return head;}else if(head.next == null && n == 1) {return null;}//获取链表长度，先遍历一次        int length = 0;//链表长度        ListNode node1 = head;        while(node1 != null) {        length++;        node1 = node1.next;        }                //再遍历第二次        ListNode node = head;        for(int i = 1; i < length-n; i++) {        node = node.next;        }        if(n == 1) {//如果删去最后一个结点        node.next = null;        }        else if(n == length) {//如果删除链表第一个结点        return head.next;        }        else {        node.next = node.next.next;        }                return head;    }//一趟完成,只遍历一次//使用两个指针，node,secNode,第一个指针比第二个快n步,当node走到链表末尾时//secNode刚好走了length-n步public ListNode removeNthFromEnd2(ListNode head, int n) {ListNode node = head;for(int i = 0; i < n; i++) {node = node.next;}if(node == null) {   //如果删除的是第一个元素return head.next;}ListNode secNode = head;while(node.next != null) {secNode = secNode.next;node = node.next;}secNode.next = secNode.next.next;return head;}}

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