83. Remove Duplicates from Sorted List II

最新推荐文章于 2023-01-13 08:10:00 发布

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最新推荐文章于 2023-01-13 08:10:00 发布

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原文链接：http://www.cnblogs.com/ycled/p/3349073.html

本文提供了一种解决方案，用于从已排序的链表中移除所有具有重复数值的节点，仅保留原始列表中的唯一数值。

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) {
        
        if(head == null || head.next == null)
            return head;
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev= dummy;
        ListNode cur = head;
        while(cur != null && cur.next != null){
            
            if(cur.next.val == cur.val){
                while(cur.next != null && cur.next.val == cur.val){
                    cur = cur.next;
                }
                prev.next = cur.next;
                cur = cur.next;
            }else{
                prev = cur;
                cur = cur.next;
            }
        }
        
        return dummy.next;
        
    }
}