Lintcode: Subarray Sum Closest

Given an integer array, find a subarray with sum closest to zero. Return the indexes of the first number and last number.

Have you met this question in a real interview? Yes
Example
Given [-3, 1, 1, -3, 5], return [0, 2], [1, 3], [1, 1], [2, 2] or [0, 4].

Challenge
O(nlogn) time

Analysis:

s[i+1] = nums[0]+....nums[i], also record the index i into s[i+1]. Sort array s, and the minimum difference between two consecutive element, is the the subarray.

 class Element implements Comparable<Element> {
         int index;
         int value;
         public Element(int index, int value) {
             this.index = index;
             this.value = value;
         }

         public int compareTo(Element other) {
             return this.value - other.value;
         }

         public int getIndex() {
             return index;
         }

         public int getValue() {
             return value;
         }
 } 

 public class Solution {
     /**
      * @param nums: A list of integers
      * @return: A list of integers includes the index of the first number
      *          and the index of the last number
      */

     public int[] subarraySumClosest(int[] nums) {
         // write your code here
         int[] res = new int[2];
         Element[] sums = new Element[nums.length+1];
         sums[0] = new Element(-1, 0);
         int sum = 0;
         for (int i=0; i<nums.length; i++) {
             sum += nums[i];
             sums[i+1] = new Element(i, sum);
         }
         Arrays.sort(sums);
         int minDif = Integer.MAX_VALUE;
         for (int i=1; i<sums.length; i++) {
             int dif = sums[i].getValue() - sums[i-1].getValue();
             if (dif < minDif) {
                 minDif = dif;
                 res[0] = Math.min(sums[i].getIndex(), sums[i-1].getIndex()) + 1;
                 res[1] = Math.max(sums[i].getIndex(), sums[i-1].getIndex());
             }
         }
         return res;
     }
 }