题目

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.

理解

删除单向链表倒数第n个元素

解决

快慢指针，找到倒数第n+1个元素，注意删除的是头指针的情况，以及只有一个节点的情况

增加假头结点

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode fakeHead = new ListNode(0);
        fakeHead.next = head;
        ListNode slow = fakeHead;
        ListNode fast = fakeHead;
        while(true){
            while(n-->0 && fast.next!=null){
                fast=fast.next;
            }
            if(fast.next==null){
                break;
            }
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return fakeHead.next;
    }
}

头结点单独处理

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode faster = head;
        ListNode slower = head;

        for (int i = 0; i < n; i++) {
            faster = faster.next;
        }
        if (faster == null) {
            head = head.next;
            return head;
        }
        while (faster.next != null) {
            slower = slower.next;
            faster = faster.next;
        }
        slower.next = slower.next.next;
        return head;
    }
}