Remove Nth Node From End of List - LeetCode

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.

Difficulty: Medium

思路：使用快指针和慢指针同时进行遍历，但快满指针间隔为 n ,因此当快指针的next为空时满指针到达要删除的结点的位置。

public ListNode removeNthFromEnd(ListNode head, int n) {

ListNode headNode = new ListNode(0);
headNode.next = head;
ListNode fastNode = headNode;
ListNode slowNode = headNode;
while(fastNode.next != null){
if(n <= 0)
slowNode = slowNode.next;
fastNode = fastNode.next;
n--;
}
if(slowNode.next != null)
slowNode.next = slowNode.next.next;
return headNode.next;
}