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 // 1 2 3 3 3 3 3 3 3 3 3 4 5 6 NULL 2 // dummy cur 3 4 public ListNode removeDeplicatesII(ListNode head) { 5 if (head == null || head.next == null) { 6 return head; 7 } 8 ListNode dummy = new ListNode(0); 9 dummy.next = head; 10 ListNode cur = dummy; 11 ListNode prev = null, tail = null; 12 13 // For example : 1 -> 1 -> null, 14 // if cur.next == null, then if we use cur.next.next, then it will come out NPE. 15 // for this example, after we delete 1, then we have dummy -> null. So if we don't consider about 16 // the situation of cur.next != null , then it will come out an NPE exception. 17 while (cur.next != null && cur.next.next != null) { 18 if (cur.next.val == cur.next.next.val) { 19 prev = cur; 20 tail = getTailNode(cur.next); 21 prev.next = tail.next; 22 } else { 23 cur = cur.next; 24 } 25 } 26 return dummy.next; 27 } 28 29 // Here we used an heler function to get the tail node. 30 public ListNode getTailNode(ListNode node) { 31 while (node != null && node.next != null) { 32 if (node.val == node.next.val) { 33 node= node.next; 34 } else { 35 break; 36 } 37 } 38 return node; 39 }
shi一样的代码1:
1 public class Solution { 2 public ListNode removeDup(ListNode head) { 3 // Write your solution here 4 if (head == null) { 5 return head; 6 } 7 ListNode dummy = new ListNode(0); 8 ListNode prev = dummy; 9 dummy.next = head; 10 ListNode cur = head; 11 while (cur != null && cur.next != null) { 12 if (cur.value == cur.next.value) { 13 ListNode newHead = remove(cur); 14 prev.next = newHead; 15 cur = newHead; 16 } else { 17 prev = cur; 18 cur = cur.next; 19 } 20 } 21 return dummy.next; 22 } 23 private ListNode remove(ListNode head) { 24 if (head == null) { 25 return head; 26 } 27 while (head.next != null && head.value == head.next.value) { 28 head = head.next; 29 } 30 // head.value != head.next.value. 31 return head.next; 32 } 33 }