How would you display an array of integers as a set of ranges? (algorithm)

Question

Given an array of integers, what is the simplest way to iterate over it and figure out all the ranges it covers? for example, for an array such as:

$numbers = ar         


        

Answer 1

My solution in Java 1.5 would be:

public static List<String> getRanges(int... in) {
    List<String> result = new ArrayList<String>();
    int last = -1;
    for (int i : in) {
        if (i != (last + 1)) {
            if (!result.isEmpty()) {
                addRange(result, last);
            }
            result.add(String.valueOf(i));
        } 
        last = i;
    }
    addRange(result, last);
    return result;
}

private static void addRange(List<String> result, int last) {
    int lastPosition = result.size() - 1;
    String lastResult = result.get(lastPosition);
    if (!lastResult.equals(String.valueOf(last))) {
        result.set(lastPosition, lastResult + "-" + last);
    }
}

public static void main(String[] args) {
    List<String> ranges = getRanges(1, 3, 4, 5, 6, 8, 11, 12, 14, 15, 16);
    System.out.println(ranges);
}

which outputs:

[1, 3-6, 8, 11-12, 14-16]

Greetz, GHad

Answer 2

module Main where

ranges :: [Int] -> [[Int]]
ranges [] = []
ranges list@(x:xs) = let adj = adjacent list in
             let len = length adj in
                 if length adj == 1
                then [[x]] ++ ranges xs
                else [[x,(last adj)]] ++ ranges (drop ((length adj) - 1) xs)
    where adjacent [] = []
          adjacent (x:xs) = if (xs /= []) && (x + 1) == head xs
             then [x] ++ adjacent (xs)
             else [x]

main = do putStrLn $ show $ ranges [1,2,3,4,5,6,8,11,12,14,15,16]

-- Output: [[1,6],[8],[11,12],[14,16]]

Here's my best shot in Haskell.

Answer 3

Create a simple range type which contains start / end of range values. Add a constructor which takes only one value and sets start = end = value. Maintain a stack of range objects, work your way through a sorted copy of the array, extend the top range or add a new range as appropriate. More specifically, when the value in the array is 1 + the end value for the range object on the to of the stack, increment the end value for that range, when it's not, push a new range (with start = end = value at index in array) onto the stack.

Answer 4

startRange = array[0];    
for(i=0;i<array.length;i++)
{
   if (array[i + 1] - array[i] > 1)
   {
     endRange = array[i];
     pushRangeOntoArray(startRange,endRange);
     i++;
     startRange = array[i]
     // need to check for end of array here
   }
}

Answer 5

Assuming the list is ordered you could start at the end and keep subtracting the next one down. While the difference is 1, you're in a range. When it's not, you start a new range.

i.e

16-15 = 1

15-14 = 1

14-12 = 2, the range is 16-14 - start a new range.

Answer 6

If the array is sorted in ascending order, then the problem is easy. Define a Range structure or class, which has a beginning and an end. Then go through the array. If the current element is one more than the previous, update Range.end, otherwise create a new range with this element as Range.begin. Store the ranges to a dynamic array or a linked list. Or just print them out as you go.

If the array may not be sorted, then sort it first.