Remove Duplicates from Sorted List II

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.

Solution:

首先用hashmap计算每个元素出现的次数，然后给list加个头指针，然后遍历链表，将次数多于1的都删掉。

 

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode deleteDuplicates(ListNode head) {
14         // Start typing your Java solution below
15         // DO NOT write main() function
16         HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
17         ListNode cur = head;
18         while(cur != null){
19             if(map.containsKey(cur.val)){
20                 map.put(cur.val,map.get(cur.val) + 1);
21             }else{
22                 map.put(cur.val,1);
23             }
24             cur = cur.next;
25         }
26         ListNode header = new ListNode(-1);
27         header.next = head;
28         cur = header;
29         while(cur.next != null){
30             if(map.get(cur.next.val) > 1){
31                 cur.next = cur.next.next;
32             }
33             else{
34                 cur = cur.next;
35             }
36         }
37         return header.next;
38     }
39 }

 

posted on 2013-09-27 07:22  Step-BY-Step  阅读(135)  评论(0)  编辑  收藏  举报

导航