Remove Duplicates from Sorted List II

题目：

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

分析：

用if判断valToDelete还是prev = prev.next，然后用while删除

很难写对，因为所有的条件必须是

while (prev.next != null && prev.next.next != null)
这是因为双层while嵌套导致的...

解法：

public class Solution {
    /**
     * @param ListNode head is the head of the linked list
     * @return: ListNode head of the linked list
     */
    public static ListNode deleteDuplicates(ListNode head) {
        // write your code here
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode prev = dummy;

        while (prev.next != null && prev.next.next != null) {
            if (prev.next.val == prev.next.next.val) {
                int valToDelete = prev.next.val;
                while (prev.next != null && prev.next.val == valToDelete) {
                    prev.next = prev.next.next;
                }
            } else {
                prev = prev.next;
            }
        }
        return dummy.next;
    }
}