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

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;
}
}