javascript sort sparse array keep indexes

Question

What is the best method to sort a sparse array and keep the elements on the same indexes? For example:

a[0] = 3, 
a[1] = 2, 
a[2] = 6,
a[7] = 4,
a[8] = 5,

I would like after the sort to have

a[0] = 2, 
a[1] = 3, 
a[2] = 4, 
a[7] = 5, 
a[8] = 6.

Answer 1

Here's one approach. It copies the defined array elements to a new array and saves their indexes. It sorts the new array and then puts the sorted results back into the indexes that were previously used.

var a = [];
a[0] = 3;
a[1] = 2; 
a[2] = 6; 
a[7] = 4; 
a[8] = 5;


// sortFn is optional array sort callback function, 
// defaults to numeric sort if not passed
function sortSparseArray(arr, sortFn) {
    var tempArr = [], indexes = [];
    for (var i = 0; i < arr.length; i++) {
        // find all array elements that are not undefined
        if (arr[i] !== undefined) {
            tempArr.push(arr[i]);    // save value
            indexes.push(i);         // save index
        }
    }
    // sort values (numeric sort by default)
    if (!sortFn) {
        sortFn = function(a,b) {
            return(a - b);
        }
    }
    tempArr.sort(sortFn);
    // put sorted values back into the indexes in the original array that were used
    for (var i = 0; i < indexes.length; i++) {
        arr[indexes[i]] = tempArr[i];
    }
    return(arr);
}

Working demo: http://jsfiddle.net/jfriend00/3ank4/