Sorting the numbers is easy in Java. **Collections** utility class provides the ‘sort’ method for sorting the numbers. [Don’t confuse with **Collection** which is an Interface]. To the Collections **Sort **method you need to input numbers in List format. Though numbers present in Array also can be converted to List and then can be sorted. Java also provides another utility class ‘**Arrays’ **which has also **Sort** method. You can use either class methods for sorting on your convenience.

Collections.sort method expects List as input [array can also be send by converting to list object using Arrays.asList(arrayObj) method]

Arrays.sort method expects Array as input [List can also be send by converting to Array object using listObj.toArray() method]

**Eg1: Sorting Numbers in List **

List <Integer> numLst1 = new ArrayList<>();
numLst1.add(20);
numLst1.add(10);
numLst1.add(04);
Collections.sort(numLst1);
//Arrays.sort(numLst1.toArray());
int f = 0;
while(f< numLst1.size()){
System.out.println(numLst1.get(f));
f++;
}

**o/p:** [4 10 20]

**Eg2: Sorting Numbers in Array **

Here we are converting Array to List and then sorting.

Integer[] numLst1 = {10, 30, 05};
Arrays.sort(numLst1);
//Collections.sort(Arrays.asList(numLst1));
int f = 0;
while(f<numLst1.length()){
System.out.println(numLst1[f].intValue());
f++;
}

**o/p:** [5 10 30]

**Eg3: Sorting numbers and finding smallest number**

Now let’s get into some deep by taking a requirement which returns the smallest number from the Array which should not be 0. Means if the Array contains numbers like [12, 20, 5, 30, 0, 50] then it should return smallest number as ‘5’ (not zero).

Integer[] returnVal = {10, 5, 20, 0};
Arrays.sort(returnVal);
//Collections.sort(Arrays.asList(returnVal));
int f= 0;
int smallValue = 0;
while(f < returnVal.length){
if (returnVal[f].intValue() !=0){
smallValue = returnVal[f].intValue();
break;
}
f++;
}
System.out.println("smallValue:"+smallValue);

**o/p: 5**

**Sorting Numbers in HashSet & HashMap [not with the help of Sort method]**

This can be achieved with the help of TreeSet. From TreeSet if you try to move to HashSet again then the insertion order will be lost. So keep the sorted result in TreeMap only or store them into LinkedHashSet where the insertion order is maintained. Similarly for HashMap sorting can be achieved with the help of TreeMap.

HashSet<Integer> hs = new HashSet<>();
hs.add(10);
hs.add(6);
hs.add(20);
hs.add(12);
System.out.println("hs:"+hs);
TreeSet<Integer> ts = new TreeSet<>();
ts.addAll(hs);
System.out.println("ts:"+ts);

**o/p: **hs:[20, 6, 10, 12] ts:[6, 10, 12, 20]