How to remove duplicate elements of ordered array

labuladong, 2020-11-11 21:02:21
remove duplicate elements ordered array

After reading this article , You can go and get the following questions :

26. Delete duplicates in sort array

83. Delete duplicate elements from the sort list


We know that for arrays , Insert at tail 、 It is more efficient to delete elements , The time complexity is O(1), But if you insert it in the middle or at the beginning 、 Remove elements , It's about moving data , The time complexity is O(N), Low efficiency .

So for the general algorithm to deal with arrays , We should only operate on the elements at the end of the array as much as possible , To avoid the extra time complexity .

This article talks about how to de duplicate an ordered array , Let's look at the title first :

obviously , Because the array has been sorted , So the repeating elements must be connected , It's not hard to find them , But if every duplicate element is found, delete it immediately , Is to delete in the middle of the array , The whole time complexity is going to be O(N^2). And the title requires us to revise it in place , In other words, auxiliary arrays can't be used , Space complexity has to be O(1).

PS: I've seriously written about 100 Multiple original articles , Hand brush 200 Daoli is the subject , All published in labuladong A copy of the algorithm , Continuous updating . Recommended collection , Write the title in the order of my article , Master all kinds of algorithm set, then put into the sea of questions, like fish .

Actually , For array related algorithms , There's a general technique : Try to avoid deleting elements in the middle , Then I'll try to change this element to the last . In this case , Finally, the elements to be deleted are dragged at the end of the array , One by one pop Just drop it , The time complexity of each operation is reduced to O(1) 了 .

According to this idea , It can also derive a common way to solve similar requirements : Double pointer technique . To be specific , It should be a slow and fast pointer .

Let's slow the pointer slow Go back to the left , Quick pointer fast Go ahead and find the way , Find an element that doesn't repeat and tell slow And let slow Take a step forward . So when fast Pointer traverses the entire array nums after ,nums[0..slow] It's not repeating elements , All the elements after that are repeating elements .

int removeDuplicates(int[] nums) {
int n = nums.length;
if (n == 0) return 0;
int slow = 0, fast = 1;
while (fast < n) {
if (nums[fast] != nums[slow]) {
// maintain nums[0..slow] No repetition
nums[slow] = nums[fast];
// The length is the index + 1
return slow + 1;

Take a look at the process of algorithm execution :

Just expand it a little bit , If I give you an ordered list , How to remove heavy ? Actually as like as two peas , The only difference is that it turns an array assignment into an operation pointer :

ListNode deleteDuplicates(ListNode head) {
if (head == null) return null;
ListNode slow = head, fast =;
while (fast != null) {
if (fast.val != slow.val) {
// nums[slow] = nums[fast]; = fast;
// slow++;
slow =;
// fast++
fast =;
// Disconnect from the following repeating elements = null;
return head;


my Online e-books Yes 100 Original articles , Hands with brushes 200 Daoli is the subject , Recommended collection ! Corresponding GitHub Algorithm Repository We've got it 70k star, Welcome to mark star !


  1. [front end -- JavaScript] knowledge point (IV) -- memory leakage in the project (I)
  2. This mechanism in JS
  3. Vue 3.0 source code learning 1 --- rendering process of components
  4. Learning the realization of canvas and simple drawing
  5. gin里获取http请求过来的参数
  6. vue3的新特性
  7. Get the parameters from HTTP request in gin
  8. New features of vue3
  9. vue-cli 引入腾讯地图(最新 api,rocketmq原理面试
  10. Vue 学习笔记(3,免费Java高级工程师学习资源
  11. Vue 学习笔记(2,Java编程视频教程
  12. Vue cli introduces Tencent maps (the latest API, rocketmq)
  13. Vue learning notes (3, free Java senior engineer learning resources)
  14. Vue learning notes (2, Java programming video tutorial)
  15. 【Vue】—props属性
  16. 【Vue】—创建组件
  17. [Vue] - props attribute
  18. [Vue] - create component
  19. 浅谈vue响应式原理及发布订阅模式和观察者模式
  20. On Vue responsive principle, publish subscribe mode and observer mode
  21. 浅谈vue响应式原理及发布订阅模式和观察者模式
  22. On Vue responsive principle, publish subscribe mode and observer mode
  23. Xiaobai can understand it. It only takes 4 steps to solve the problem of Vue keep alive cache component
  24. Publish, subscribe and observer of design patterns
  25. Summary of common content added in ES6 + (II)
  26. No.8 Vue element admin learning (III) vuex learning and login method analysis
  27. Write a mini webpack project construction tool
  28. Shopping cart (front-end static page preparation)
  29. Introduction to the fluent platform
  30. Webpack5 cache
  31. The difference between drop-down box select option and datalist
  32. CSS review (III)
  33. Node.js学习笔记【七】
  34. Node.js learning notes [VII]
  35. Vue Router根据后台数据加载不同的组件(思考-&gt;实现-&gt;不止于实现)
  36. Vue router loads different components according to background data (thinking - & gt; Implementation - & gt; (more than implementation)
  37. 【JQuery框架,Java编程教程视频下载
  38. [jQuery framework, Java programming tutorial video download
  39. Vue Router根据后台数据加载不同的组件(思考-&gt;实现-&gt;不止于实现)
  40. Vue router loads different components according to background data (thinking - & gt; Implementation - & gt; (more than implementation)
  41. 【Vue,阿里P8大佬亲自教你
  42. 【Vue基础知识总结 5,字节跳动算法工程师面试经验
  43. [Vue, Ali P8 teaches you personally
  44. [Vue basic knowledge summary 5. Interview experience of byte beating Algorithm Engineer
  45. 【问题记录】- 谷歌浏览器 Html生成PDF
  46. [problem record] - PDF generated by Google browser HTML
  47. 【问题记录】- 谷歌浏览器 Html生成PDF
  48. [problem record] - PDF generated by Google browser HTML
  49. 【JavaScript】查漏补缺 —数组中reduce()方法
  50. [JavaScript] leak checking and defect filling - reduce() method in array
  51. 【重识 HTML (3),350道Java面试真题分享
  52. 【重识 HTML (2),Java并发编程必会的多线程你竟然还不会
  53. 【重识 HTML (1),二本Java小菜鸟4面字节跳动被秒成渣渣
  54. [re recognize HTML (3) and share 350 real Java interview questions
  55. [re recognize HTML (2). Multithreading is a must for Java Concurrent Programming. How dare you not
  56. [re recognize HTML (1), two Java rookies' 4-sided bytes beat and become slag in seconds
  57. 【重识 HTML ,nginx面试题阿里
  58. 【重识 HTML (4),ELK原来这么简单
  59. [re recognize HTML, nginx interview questions]
  60. [re recognize HTML (4). Elk is so simple