Subarray Sum Closest

Question

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

Given [-3, 1, 1, -3, 5], return [0, 2], [1, 3],[1, 1], [2, 2] or [0, 4].

Answer

这道题延续Subarray Sum的思路，即将[0, i]的sum存起来。这里构造了一个新的类，Pair。一方面存sum值，一方面存index。然后根据sum值排序，算相邻sum值的差值。差值最小的即为结果。时间复杂度是O(nlogn)，空间复杂度是O(n)。

注意：假设[0-2]和[0-4]的sum值的差值最小，那么结果为index[3,4]

public class Solution {
    
    class Pair {
        public int sum;
        public int index;
        public Pair(int sum, int index) {
            this.sum = sum;
            this.index = index;
        }
    }
    /**
     * @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[] result = new int[2];
        if (nums == null || nums.length == 0) {
            return result;
        }
        int len = nums.length;
        int sum = 0;
        List<Pair> list = new ArrayList<Pair>();
        for (int i = 0; i < len; i++) {
            sum += nums[i];
            Pair tmp = new Pair(sum, i);
            list.add(tmp);
        }
        Collections.sort(list, new Comparator<Pair>() {
            public int compare(Pair a, Pair b) {
                return (a.sum - b.sum);
            }
        });
        int min = Integer.MAX_VALUE;
        for (int i = 1; i < len; i++) {
            int diff = list.get(i).sum - list.get(i - 1).sum;
            if (diff < min) {
                min = diff;
                int index1 = list.get(i).index;
                int index2 = list.get(i - 1).index;
                if (index1 < index2) {
                    result[0] = index1 + 1;
                    result[1] = index2;
                } else {
                    result[0] = index2 + 1;
                    result[1] = index1;
                }
            }
        }
        return result;
    }
}