zoukankan html css js c++ java

[LeetCode] 82. Remove Duplicates from Sorted List II

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Example 1:

Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]

Example 2:

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

删除排序链表中的重复元素II。

题意跟版本一很接近，唯一的区别是，版本一是请你删除重复元素，使得output里面每个元素只出现一次。版本二是请你删除所有出现超过一次的元素。思路是需要创建一个dummy节点，放在所有需要遍历的节点之前，遍历的时候，找是否有两个节点的val相同，找到后记下这个val。再往后遍历的时候，只要遇到这个val就跳过。

比如第一个例子好了，当cur遍历到2的时候，cur.next == 3, cur.next.next == 3；此时记录sameVal = cur.next.val。接着从cur.next开始判断，只要cur.next.val == 3的时候，就cur.next = cur.next.next。

如果是比如一开始遍历到1的时候好了，此时cur.next == 2，cur.next.next == 3，两者并不相等，所以直接就遍历下一个节点，cur = cur.next。

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {ListNode} head
 3  * @return {ListNode}
 4  */
 5 var deleteDuplicates = function(head) {
 6     // corner case
 7     if (head === null || head.next === null) return head;
 8 
 9     // normal case
10     let dummy = new ListNode(0);
11     dummy.next = head;
12     let cur = dummy;
13     while (cur.next !== null && cur.next.next !== null) {
14         if (cur.next.val === cur.next.next.val) {
15             let sameVal = cur.next.val;
16             while (cur.next !== null && cur.next.val === sameVal) {
17                 cur.next = cur.next.next;
18             }
19         } else {
20             cur = cur.next;
21         }
22     }
23     return dummy.next;
24 };

Java实现

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode deleteDuplicates(ListNode head) {
11         // corner case
12         if (head == null || head.next == null) {
13             return head;
14         }
15 
16         // normal case
17         ListNode dummy = new ListNode(0);
18         dummy.next = head;
19         // 注意这里cur是从dummy开始
20         ListNode cur = dummy;
21         while (cur.next != null && cur.next.next != null) {
22             if (cur.next.val == cur.next.next.val) {
23                 int sameVal = cur.next.val;
24                 while (cur.next != null && cur.next.val == sameVal) {
25                     cur.next = cur.next.next;
26                 }
27             } else {
28                 cur = cur.next;
29             }
30         }
31         return dummy.next;
32     }
33 }