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;
    }
}