Converting an int array to a String array
So I have this "list" of ints. It could be a Vector
, int[]
, List<Integer>
, whatever.
My goal though is to sort the ints and end up with a String[]
. How the int array starts out as is up in the air.
ex:
Start with:{5,1,2,11,3}
End with: String[] = {"1","2","3","5","11"}
Is 开发者_高级运维there anyway to do this without a for loop? I have a for loop now for collecting the ints. I would rather skip doing another for loop.
int[] nums = {5,1,2,11,3}; //List or Vector
Arrays.sort(nums); //Collections.sort() for List,Vector
String a=Arrays.toString(nums); //toString the List or Vector
String ar[]=a.substring(1,a.length()-1).split(", ");
System.out.println(Arrays.toString(ar));
UPDATE:
A shorter version:
int[] nums = {-5,1,2,11,3};
Arrays.sort(nums);
String[] a=Arrays.toString(nums).split("[\\[\\]]")[1].split(", ");
System.out.println(Arrays.toString(a));
Use a Stream
which is available from Java 8. To get a Stream
instance with "list" of ints:
- For
int[]
IntStream intStream = Arrays.Stream(nums);
orStream<Integer> intStream = Arrays.Stream(nums).boxed();
if you need the same class as bottom one.
- For any classes with
Collection<Integer>
interface (ex.Vector<Integer>
,List<Integer>
)Stream<Integer> intStream = nums.stream();
Finally, to get a String[]
:
String[] answer = intStream.sorted().mapToObj(String::valueOf).toArray(String[]::new);
Can I use a while
loop instead?
@Test
public void test() {
int[] nums = {5,1,2,11,3};
Arrays.sort(nums);
String[] stringNums = new String[nums.length];
int i = 0;
while (i < nums.length) {
stringNums[i] = String.valueOf(nums[i++]);
}
Assert.assertArrayEquals(new String[]{"1","2","3","5","11"}, stringNums);
}
Using JUnit assertions.
Sorry, I'm being flippant. But saying you can't use a for
loop is daft - you've got to iterate over the list somehow. If you're going to call a library method to sort it for you (cf Collections.sort()) - that will be looping somehow over the elements.
Simple solution using Guava:
public List<String> toSortedStrings(List<Integer> ints) {
Collections.sort(ints);
return Lists.newArrayList(Iterables.transform(ints,
Functions.toStringFunction()));
}
Obviously, this solution (like any other) is going to use loops internally, but it gets it out of the code you have to read. You could also avoid changing the order in ints
by passing the result of Ordering.natural().sortedCopy(ints)
to transform
instead of using Collections.sort
first. Also, the Lists.newArrayList
part is not necessary if you don't need to be able to add new elements to the resulting list.
The shortened version of that method body, with static imports:
return transform(Ordering.natural().sortedCopy(ints), toStringFunction());
If you use a TreeSet, I have a (longish) one-liner for you (assuming items
is the TreeSet):
final String[] arr =
items.toString() // string representation
.replaceAll("\\D+", " ") // replace all non digits with spaces
.trim() // trim ends
.split(" "); // split by spaces
Test code:
Set<Integer> items = new TreeSet<Integer>(Arrays.asList(5, 1, 2, 11, 3));
// insert above code here
System.out.println(Arrays.toString(arr));
Output:
[1, 2, 3, 5, 11]
EDIT:
OK, here is a different version that works with the int array directly. But unfortunately it's not a one-liner. However, it does keep duplicates and it's probably faster
EDIT again:
Bug fixed and negative numbers supported, as requested:
EDIT once more: only one regex pass and no trim
final int[] in = { 5, 1, 2, 11, 3, 2, -5 }; // with duplicate
Arrays.sort(in);
final String[] out =
Arrays.toString(in)
.replaceAll("(?:\\[?)([-\\d]+)(?:\\]?)", "$1") // just remove [ and ]
.split("\\s*,\\s*"); // split by comma
System.out.println(Arrays.toString(out));
Output:
[-5, 1, 2, 2, 3, 5, 11]
Or completely without regex (apart from split()), but with one more step added:
final int[] in = { 5, 1, 2, 11, 3, 2, -5 }; // with duplicate
Arrays.sort(in);
final String stringRep = Arrays.toString(in);
final String[] out =
stringRep.substring(1, stringRep.length() - 1).split("\\s*,\\s*");
System.out.println(Arrays.toString(out));
Output:
[-5, 1, 2, 2, 3, 5, 11]
Update: stripped whitespace from my last two solutions, hope you're happy now :-)
I would rather skip doing another for loop.
That's silly. It's a silly desire and a silly basis for undertaking a code exercise. If you can better express the qualities that you want your code to have, then we've got something to talk about - that it should be easy to read, say, or performant, or testable, or robust. But "I'd rather skip it" just doesn't give us anything useful to work with.
Using Functional Java,
import fj.data.List;
import static fj.data.List.*;
import static fj.pre.Show.*;
.
.
.
final List<Integer> xs = list(5,1,2,11,3);
final List<String> ys = xs.sort(Ord.intOrd).map(
new F<Integer, String>() {
@Override public String f(final Integer i) {
return String.valueOf(i);
}
}
);
listShow(stringShow).println(ys);
How about something like this:
List<String> stringList = new ArrayList<String>();
List<Integer> list = new ArrayList<Integer>(Arrays.asList(5,1,2,11,3));
Collections.sort(list);
Iterator<Integer> iterator = list.iterator();
while(iterator.hasNext()) {
stringList.add(iterator.next().toString());
}
System.out.println(stringList);
Using Eclipse Collections MutableIntList
:
String[] result = IntLists.mutable.with(5, 1, 2, 11, 3)
.sortThis()
.collect(Integer::toString)
.toArray(new String[]{});
Assert.assertArrayEquals(
new String[]{"1", "2", "3", "5", "11"}, result);
Or trading some readability for potential efficiency:
MutableIntList intList = IntLists.mutable.with(5, 1, 2, 11, 3).sortThis();
String[] result = intList.injectIntoWithIndex(
new String[intList.size()],
(r, each, index) -> {
r[index] = Integer.toString(each);
return r;
});
Assert.assertArrayEquals(
new String[]{"1", "2", "3", "5", "11"}, result);
Note: I am a committer for Eclipse Collections
We can solve this problem using regular expression as follows:
int[] intArray = {1, 2, 3, 4, 5, 6, 7, 8};
String str = new String(Arrays.toString(intArray));
String stripNoneDigits= str.replaceAll("[^\\d]", "");
String[] stringArray = stripNoneDigits.split("");
You can use Collections.sort()
and then iterate over the list and collectString.valueOf()
each element.
http://download.oracle.com/javase/6/docs/api/java/util/Collections.html#sort%28java.util.List%29
For a Vector, you would first get a List with Collections.list(Enumeration e)
.
For an array, you would use Arrays.sort()
instead of Collections.sort()
.
http://download.oracle.com/javase/6/docs/api/java/util/Arrays.html#sort%28int%5b%5d%29
Why don't you simply cast those values to String within the original for loop, creating a String array rather than an int array? Assuming that you're gathering your initial integer from a starting point and adding to it on each for loop iteration, the following is a simple methodology to create a String array rather than an int array. If you need both int and String arrays with the same values in them, create them both in the same for loop and be done with it.
yourInt = someNumber;
for (int a = 0; a < aLimit; a ++) {
String stringName = String.valueOf(yourInt);
StringArrayName[a] = stringName;
yourInt ++;
}
Or, if you need both:
yourInt = someNumber;
for (int a = 0; a < aLimit; a ++) {
String stringName = String.valueOf(yourInt);
StringArrayName[a] = stringName;
intArrayName[a] = yourInt;
yourInt ++;
}
I agree with everyone else. For loops are easy to construct, require almost no overhead to run, and are easy to follow when reading code. Elegance in simplicity!
Java 8 way, no loops at all:
// Given
int[] array = {-5, 8, 3, 10, 25};
// When
String[] actual = Arrays.stream(array)
.sorted()
.mapToObj(String::valueOf)
.toArray(String[]::new);
// Then
String[] expected = {"-5", "3", "8", "10", "25"};
assertArrayEquals(expected, actual);
Arrays.sort(nums); var stringArray = (nums.toString()).split(',').map(String);
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