LeetCode——Remove Duplicates from Sorted List II
2023-09-11 14:14:43 时间
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:给定一个已排序的链表,删除反复的元素。仅仅留下源链表中唯一出现的那些元素。
思路:首先防止把头节点删除。构建一个虚拟的头节点。
因为是已排序的。相等的元素就在相邻的位置上。发现有相邻的元素。则指针依次向后指,直到没有相等的相邻元素为止。
public ListNode deleteDuplicates(ListNode head) { if (head == null || head.next == null) return head; ListNode dummy = new ListNode(0); dummy.next = head; head = dummy; while (head.next != null && head.next.next != null) { if (head.next.val == head.next.next.val) { int val = head.next.val; while (head.next != null && head.next.val == val) { head.next = head.next.next; } } else { head = head.next; } } return dummy.next; } // Definition for singly-linked list. public class ListNode { int val; ListNode next; ListNode(int x) { val = x; next = null; } }
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