LeetCode 082~083 Remove Duplicates from Sorted List

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本文介绍如何从已排序的链表中移除重复元素，包括只保留每个元素一次的情况及完全移除所有重复元素的方法。通过使用集合数据结构和迭代方式重新构建链表，确保链表中不再包含重复项。

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Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

题目解析：去除重复的，只显示单个的内容

代码：

/**
 * 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) {
		Set<Integer> set=new TreeSet<Integer>();
		while(head!=null){
			if(!set.contains(head.val)){
				set.add(head.val);
			}
			head=head.next;
		}
		ListNode result,temp;
		Iterator<Integer> it=set.iterator();
		if(it.hasNext()){
			result=new ListNode(it.next());
			temp=result;
			while(it.hasNext()){
				temp.next=new ListNode(it.next());
				temp=temp.next;
			}
			return result;
		}
		return null;

	}
}

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.

题目解析：在上一题的基础上，把重复的部分去掉，只显示没有重复的

代码：

/**
 * 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) {
		Set<Integer> set=new TreeSet<Integer>();
		Set<Integer> set0=new LinkedHashSet<Integer>();
		while(head!=null){
			if(!set.contains(head.val)){
				set.add(head.val);
				set0.add(head.val);
			}else{
				set0.remove(head.val);
			}
			head=head.next;
		}
		ListNode result,temp;
		Iterator<Integer> it=set0.iterator();
		if(it.hasNext()){
			result=new ListNode(it.next());
			temp=result;
			while(it.hasNext()){
				temp.next=new ListNode(it.next());
				temp=temp.next;
			}
			return result;
		}
		return null;

	}
}