Leetcode 83. Remove Duplicates from Sorted List-从有序链表中删除重复节点，重复的节点只保留一个

Posted 2022-04-05 二十六画生的博客

Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

Example 1:

Input: head = [1,1,2]
Output: [1,2]

Example 2:

Input: head = [1,1,2,3,3]
Output: [1,2,3]

Constraints:

• The number of nodes in the list is in the range [0, 300].
• -100 <= Node.val <= 100
• The list is guaranteed to be sorted in ascending order.

package com.linkedlist;

/**
 * @Author you guess
 * @Date 2022/4/1 17:17
 * @Version 1.0
 * @Desc 删除重复节点（如果有2个节点值相同，只保留1个节点；重复的节点只保留一个）
 */
public class Leetcode_83_RemoveDuplicatesfromSortedList 


    public static void main(String[] args) 
        ListNode node72 = new ListNode(7, null);
        ListNode node7 = new ListNode(7, node72);
        ListNode node6 = new ListNode(6, node7);
        ListNode node5 = new ListNode(5, node6);
        ListNode node42 = new ListNode(4, node5);
        ListNode node41 = new ListNode(4, node42);
        ListNode node40 = new ListNode(4, node41);
        ListNode node3 = new ListNode(3, node40);
        ListNode node22 = new ListNode(2, node3);
        ListNode node21 = new ListNode(2, node22);
        ListNode node20 = new ListNode(2, node21);
        ListNode node11 = new ListNode(1, node20);
        ListNode node1 = new ListNode(1, node11);
        Leetcode_83_RemoveDuplicatesfromSortedList main = new Leetcode_83_RemoveDuplicatesfromSortedList();
        main.printListNode(main.deleteDuplicates(node1));
    


    /**
     * Runtime: 0 ms, faster than 100.00% of Java online submissions for Remove Duplicates from Sorted List.
     * Memory Usage: 41.6 MB, less than 94.40% of Java online submissions for Remove Duplicates from Sorted List.
     * <p>
     * 每次:head与head.next值相等时，pre.next = head.next
     * head与head.next值不相等时，pre=最后一个重复值，head=下一个值
     *
     * @param head
     * @return
     */
    public ListNode deleteDuplicates(ListNode head) 
        ListNode sentinel = new ListNode(0, head);
        ListNode pre = sentinel;
        while (head != null) 
            if (head.next != null && head.val == head.next.val) 
                pre.next = head.next;
                head = head.next;
             else 
                pre = pre.next; //pre指向的链表也是连续的
                head = head.next;
            
        
        return sentinel.next;
    


    public void printListNode(ListNode head) 
        while (head != null) 
            System.out.println(head.val);
            head = head.next;