leetcode - Contains Duplicate

Problem:
Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.

Solution:
在解决问题之前先巩固几个知识点：
（1）set、map和list：
set：无序不重复 .add
map：键值对 .put
list：有序可重复

（2）int和Integer的区别
原始类型 引用类型(封装类型)
int 是基本类型，直接存数值，而integer是对象，用一个引用指向这个对象

（3）一定记得做预处理：在这里举个例子，当你拿到一个数组，你先要判断这个数组是不是空的，可以通过长度是否为0判断
我的思路：将数组放在set中，而set是不允许有重复值的。如果set的size和数组的length相等，说明无重复。

public class Solution {    public boolean containsDuplicate(int[] nums) {        if(nums.length == 0) return false;        HashSet<Integer> hs = new HashSet<Integer>();        for(int i = 0 ; i < nums.length ; i++){            int one = nums[i];            hs.add(one);        }        if(hs.size() == nums.length){            return false;        }else{            return true;        }    }}

II-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.

solution:

public class Solution {    public boolean containsNearbyDuplicate(int[] nums, int k) {        HashMap<Integer,Integer> hm = new HashMap<Integer,Integer>();        if (nums.length == 0)  return false;        for(int i=0 ; i < nums.length ;  i++){            if(!hm.containsKey(nums[i])){                hm.put(nums[i],i);            }else{                int before = hm.get(nums[i]);                int gap = i - before;                if(gap <= k){                    return true;                }else{                    hm.put(nums[i],i);                }            }        }        return false;    }}