未完成--Sum of Pairs

Sum of Pairs
Given a list of integers and a single sum value, return the first two values (parse from the left please) in order of appearance that add up to form the sum.
sum_pairs([11, 3, 7, 5],         10)
#              ^--^      3 + 7 = 10
== [3, 7]
sum_pairs([4, 3, 2, 3, 4],         6)
#          ^-----^         4 + 2 = 6, indices: 0, 2 *
#             ^-----^      3 + 3 = 6, indices: 1, 3
#                ^-----^   2 + 4 = 6, indices: 2, 4
#  * entire pair is earlier, and therefore is the correct answer
== [4, 2]
sum_pairs([0, 0, -2, 3], 2)
#  there are no pairs of values that can be added to produce 2.
== None/nil/undefined (Based on the language)
sum_pairs([10, 5, 2, 3, 7, 5],         10)
#              ^-----------^   5 + 5 = 10, indices: 1, 5
#                    ^--^      3 + 7 = 10, indices: 3, 4 *
#  * entire pair is earlier, and therefore is the correct answer
== [3, 7]
Negative numbers and duplicate numbers can and will appear.
NOTE: There will also be lists tested of lengths upwards of 10,000,000 elements. Be sure your code doesn't time out.

var sum_pairs = function (ints, s) {
    for (var i = 0, l = ints.length, r = []; i < l; i++) {
        for (var j = 1; j < l; j++) {
            if (ints[i] + ints[j] == s) {
                if (r[1]==null||j + i >= r[0] + r[1]) {
                    r[0] = i;
                    r[1] = j;
                }
            }
        }
    }
    return r[0]?[ints[r[0]], ints[r[1]]]:"undefined for sum = " + s;
}