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[LeetCode] 83. Remove Duplicates from Sorted List

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.

删除排序链表中的重复元素。

思路很简单，每次去看一下当前节点和下一个节点的val是否相同，如是，则跳过下一个节点。

时间O(n)

空间O(1)

注意1 - 1 - 1这样的case怎么处理。当满足if的条件的时候，一定要写else。

JavaScript实现

 1 /**
 2  * Definition for singly-linked list.
 3  * function ListNode(val) {
 4  *     this.val = val;
 5  *     this.next = null;
 6  * }
 7  */
 8 /**
 9  * @param {ListNode} head
10  * @return {ListNode}
11  */
12 var deleteDuplicates = function(head) {
13     // corner case
14     if (head === null || head.next === null) {
15         return head;
16     }
17 
18     // normal case
19     let cur = head;
20     while (cur !== null && cur.next !== null) {
21         if (cur.val === cur.next.val) {
22             cur.next = cur.next.next;
23         } else {
24             cur = cur.next;
25         }
26     }
27     return head;
28 };

Java实现

 1 class Solution {
 2     public ListNode deleteDuplicates(ListNode head) {
 3         // corner case
 4         if (head == null || head.next == null) {
 5             return head;
 6         }
 7 
 8         // normal case
 9         ListNode cur = head;
10         while (cur.next != null) {
11             if (cur.next.val == cur.val) {
12                 cur.next = cur.next.next;
13             } else {
14                 cur = cur.next;
15             }
16         }
17         return head;
18     }
19 }