LeetCode 82 Remove Duplicates from Sorted List II

Here, I used a helper function to help me get the tail node of duplicate number list.

Store the prev and tail node.

cur.next != null && cur.next.next != null.

简洁版代码：

//          1    2    3    3    3    3    3    3    3    3    3    4    5    6 NULL
// dummy        cur

public ListNode removeDeplicatesII(ListNode head) {
if (head == null || head.next == null) {
return head;
}
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode cur = dummy;
        ListNode prev = null, tail = null;
        
  // For example : 1 -> 1 -> null, 
  // if cur.next == null, then if we use cur.next.next, then it will come out NPE.
  // for this example, after we delete 1, then we have dummy -> null. So if we don't consider about 
  // the situation of cur.next != null , then it will come out an NPE exception.
        while (cur.next != null && cur.next.next != null) {
            if (cur.next.val == cur.next.next.val) {
                prev = cur;
                tail = getTailNode(cur.next);
                prev.next = tail.next;
            } else {
                cur = cur.next;
            }
        }
        return dummy.next;
    }

    // Here we used an heler function to get the tail node.
    public ListNode getTailNode(ListNode node) {
        while (node  != null && node.next != null) {
            if (node.val == node.next.val) {
                node= node.next;
            } else {
                break;
            }
        }
        return node;
    }

shi一样的代码1:

public class Solution {
  public ListNode removeDup(ListNode head) {
    // Write your solution here
    if (head == null) {
         return head; 
    }
    ListNode dummy = new ListNode(0);
    ListNode prev = dummy;
    dummy.next = head;
    ListNode cur = head;
    while (cur != null && cur.next != null) {
         if (cur.value == cur.next.value) {
           ListNode newHead = remove(cur);
        prev.next = newHead;
        cur = newHead;
      } else {
        prev = cur;
        cur = cur.next;
      }
    }
    return dummy.next;
  }
  private ListNode remove(ListNode head) {
    if (head == null) {
     return head; 
    }
    while (head.next != null && head.value == head.next.value) {
          head = head.next;
    }
    // head.value != head.next.value.
    return head.next;
  }
}