remove duplicates from sorted list, remove nth node

Linkedlist中的remove，记得要构造fakehead！

1. Remove nth node from the end of list

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

2. remove duplicates from sorted list I

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode pre = head;
        ListNode cur = head.next;
        while (cur != null) {
            if (pre.val == cur.val) {
                cur = cur.next;
            } else {
                pre.next = cur;
                pre = cur;
                cur = cur.next;
            }
        }
        pre.next = null;
        return head;
    }
}

3, remove duplicates from sorted list II

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode fakehead = new ListNode(0);
        fakehead.next = head;
        ListNode node = fakehead;
        while (node.next != null && node.next.next != null) {
            if (node.next.val == node.next.next.val) {
                ListNode temp = node.next.next;
                while (temp.next != null && temp.next.val == temp.val) {
                    temp = temp.next;
                }
                node.next = temp.next;
            } else {
                node = node.next;
            }
        }
        return fakehead.next;
    }
}