1 ms Java Solution  --> O(1) Space Complexity  ,O(N) Time Complexity
STEPS :\n Make two arrays of 100 length to keep the track of the count of ListNodes Value.\n Iterate the Linked List for counting .\n Create a dummy node and ma
***STEPS :***\n* Make two arrays of 100 length to keep the track of the count of ListNodes Value.\n* Iterate the Linked List for counting .\n* Create a dummy node and marks its next to head node.\n* Create a temp node to iterate the linked list .\n* Now while iterating check if nexts nodes count in the array is 1 or not if it is one then move to next node else move to next of next node .\n* Return dummy\'s next node\n\n\n\n**TIME COMPLEXITY ANALYSIS:**\n* *TIME COMPLEXITY :* O(N) {For counting} + O(N) {For iterating the array} =O(2N)\n* *SPACE COMPLEXITY :* O(1) {For keep track of count of Nodes value = O(1)\n\n\n*Code for more understanding*\n```\nclass Solution {\n    public ListNode deleteDuplicates(ListNode head) {\n        int pos [] = new int [101];\n        int neg [] = new int [100];\n        ListNode temp = head;\n        while(temp!=null){\n            if(temp.val>=0){\n                pos[temp.val]++;\n            }\n            else{\n                neg[Math.abs(temp.val)]++;\n            }\n            temp = temp.next;\n        }\n        ListNode dummy = new ListNode(-1);\n        dummy .next = head;\n        temp = dummy;\n        while(temp.next!=null){\n            \n            if(temp.next.val>=0){\n                if(pos[temp.next.val]!=1){\n                    temp.next = temp.next.next;\n                }\n                else temp = temp.next;\n            }\n            else{\n               if(neg[Math.abs(temp.next.val)]!=1){\n                   temp.next = temp.next.next;\n               } \n                else temp = temp.next;\n            }\n        }\n        return dummy.next;\n    }\n}\n```
['Linked List', 'Iterator', 'Java']