[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。
给定一个已排序的链表的头 head , 删除所有重复的元素,使每个元素只出现一次 。返回 已排序的链表 。
思路
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。
复杂度
时间O(n)
空间O(1)
代码
Java实现
/**
* 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) {
// corner case
if (head == null || head.next == null) {
return head;
}
// normal case
ListNode cur = head;
while (cur.next != null) {
if (cur.val == cur.next.val) {
cur.next = cur.next.next;
} else {
cur = cur.next;
}
}
return head;
}
}
JavaScript实现
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var deleteDuplicates = function(head) {
// corner case
if (head === null || head.next === null) {
return head;
}
// normal case
let cur = head;
while (cur !== null && cur.next !== null) {
if (cur.val === cur.next.val) {
cur.next = cur.next.next;
} else {
cur = cur.next;
}
}
return head;
};
相关题目
83. Remove Duplicates from Sorted List
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