Created
May 7, 2013 08:27
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Remove Duplicates from Sorted List II
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/* | |
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. | |
For example, | |
Given 1->2->3->3->4->4->5, return 1->2->5. | |
Given 1->1->1->2->3, return 2->3. | |
*/ | |
/** | |
* Definition for singly-linked list. | |
* public class ListNode { | |
* int val; | |
* ListNode next; | |
* ListNode(int x) { | |
* val = x; | |
* next = null; | |
* } | |
* } | |
*/ | |
public class Solution { | |
public ListNode deleteDuplicates(ListNode head) { | |
// Start typing your Java solution below | |
// DO NOT write main() function | |
if(head == null) return null; | |
HashMap<Integer, Integer> set = new HashMap<Integer, Integer>(); | |
set.put(head.val, 1); | |
ListNode cur = head.next; | |
while(cur != null) { | |
if(set.containsKey(cur.val)) { | |
set.put(cur.val, set.get(cur.val) + 1); | |
} else { | |
set.put(cur.val, 1); | |
} | |
cur = cur.next; | |
} | |
cur = head; | |
ListNode dummy = new ListNode(-1); | |
dummy.next = head; | |
ListNode prev = dummy; | |
while(cur != null) { | |
if(set.get(cur.val) > 1) { | |
prev.next = cur.next; | |
cur = cur.next; | |
} else { | |
prev = cur; | |
cur = cur.next; | |
} | |
} | |
return dummy.next; | |
} | |
} |
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