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;
     }
 }