83. Remove Duplicates from Sorted List

最新推荐文章于 2024-01-17 14:07:10 发布

xdhc304

最新推荐文章于 2024-01-17 14:07:10 发布

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分类专栏： leetcode 文章标签： leetcode-java

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本文介绍了一种算法，用于从已排序的链表中移除所有重复元素，确保每个元素只出现一次。提供了三种实现方法，包括直接遍历、使用前驱节点以及迭代更新当前节点的后续指针。

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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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null)return head;
        ListNode cur = head,start = head;
        while(start != null){
            if(cur == null){
                start.next = null;
                break;
            }
            if(cur.val == start.val)cur = cur.next;
            else {
                start.next = cur;
                start = cur;
                cur = cur.next;
            }
        }
        return head;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode prev = new ListNode(-1);
        ListNode curr = head;
        while(curr != null) {
            if(curr.val == prev.val) {
                prev.next = curr.next;
            } else {
                prev = curr;
            }
            curr = curr.next;
        }
        return head;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode curr=head;
        while(curr!=null) {
            ListNode temp=curr;
            while(temp!=null && temp.val==curr.val) {
                temp=temp.next;
            }
            curr.next=temp;
            curr=curr.next;
        }
        return head;
    }
}