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This article mainly introduces how to solve the problem of repetitive elements in leetcode. It is very detailed and has a certain reference value. Interested friends must read it!

Topic link

Https://leetcode-cn.com/problems/contains-duplicate-ii/

Topic description

Given an array of integers and an integer k, it is determined whether there are two different indexes I and j in the array, such that nums [I] = nums [j], and the absolute value of the difference between I and j is maximum k.

Example 1:

Input: nums = [1, 2, 2, 3, 1], k = 3

Output: true

Example 2:

Input: nums = [1j0pr 1pl], k = 1

Output: true

Example 3:

Input: nums = [1, 2, 3, 1, 1, 2, 3], k = 2

Output: false

The idea of solving the problem

Tag: hash

Maintain a hash table that always contains at most k elements. When a duplicate value occurs, it indicates that there are duplicate elements within the k distance.

Traversing one element at a time adds it to the hash table, and removes the first number if the size of the hash table is greater than k

Time complexity: O (n), n is the length of the array

Code

Java version

Class Solution {

Public boolean containsNearbyDuplicate (int [] nums, int k) {

HashSet set = new HashSet ()

For (int I = 0; I

< nums.length; i++) { if(set.contains(nums[i])) { return true; } set.add(nums[i]); if(set.size() >

K) {

Set.remove (Nums [I-k])

}

}

Return false

}

}

JavaScript version

/ * *

* @ param {number []} nums

* @ param {number} k

* @ return {boolean}

, /

Var containsNearbyDuplicate = function (nums, k) {

Const set = new Set ()

For (let I = 0; I

< nums.length; i++) { if(set.has(nums[i])) { return true; } set.add(nums[i]); if(set.size >

K) {

Set.delete (Nums [I-k])

}

}

Return false

}

Drawing and interpretation

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