题目:
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
Example 1:
Input: 1->2->3->3->4->4->5
Output: 1->2->5
Example 2:
Input: 1->1->1->2->3
Output: 2->3
public class RemoveDuplicatesSortedListII {
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
public ListNode deleteDuplicates(ListNode head) {
if(head==null||head.next==null)
return head;
ListNode cur = head;
ListNode dummy = new ListNode(0);
dummy.next =head;
ListNode prev= dummy;
while(cur !=null){
while(cur .next !=null && cur .val==cur .next.val){
cur =cur .next;
}
if(prev.next!=cur ){
prev.next=cur .next;
}else{
prev=prev.next;
}
cur =cur .next;
}
return dummy.next;
}
}
本文介绍了一种在已排序链表中移除所有具有重复数值节点的算法,仅保留原始列表中的唯一数值。通过示例展示输入为1->2->3->3->4->4->5时,输出为1->2->5;当输入为1->1->1->2->3时,输出为2->3。文章提供了一个名为RemoveDuplicatesSortedListII的Java类实现。
684

被折叠的 条评论
为什么被折叠?



