[LeetCode-interview algorithm classic-Java implementation] [219-Contains Duplicate II (including Duplicate element II)], leetcode -- java

[LeetCode-interview algorithm classic-Java implementation] [219-Contains Duplicate II (including Duplicate element II)], leetcode -- java
[219-Contains Duplicate II (including Duplicate element II )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Download the Code [https://github.com/wang-jun-chao]Original question

Given an array of integers and an integer k, find out whether there are two distinct indices I and j in the array such that nums [I] = nums [j] and the difference between I and j is at most k.

Theme

Given an integer array nums and an integer k, if two different subscripts I and j meet the requirements of nums [I] = nums [j] and | I-j | <= k, true is returned. Otherwise, false is returned.

Solutions

For nums [0... N-1], stored in a map, (muns [I], I), if the key nums [k] already exists, compare the difference between the previous subscript and the current lower mark, if the difference value is not greater than k, it indicates the two values that meet the condition. Otherwise, the new subscript is used as the value.

Code Implementation

Algorithm Implementation class

Import java. util. hashMap; import java. util. map; public class Solution {public boolean containsNearbyDuplicate (int [] nums, int k) {// if (nums = null | nums. length <2 | k <1) {return false;} Map <Integer, Integer> map = new HashMap <> (); for (int I = 0; I <nums. length; I ++) {// if no corresponding key is added to if (! Map. containsKey (nums [I]) {map. put (nums [I], I) ;}// there is already a subscript corresponding to the key-value Pair else {// The original saved value, it must be smaller than the current subscript int value = map. get (nums [I]); if (I-value <= k) {return true;} map. put (nums [I], I) ;}} return false ;}}

Evaluation Result

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