Subarray Sum Closest

最新推荐文章于 2019-05-02 19:55:10 发布

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最新推荐文章于 2019-05-02 19:55:10 发布

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本文介绍了一种高效算法，用于从整数数组中找到和最接近零的子数组，并返回该子数组的起始和结束索引。通过定义一个自定义元素类并使用排序技巧，可以在O(n log n)的时间复杂度内解决此问题。

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

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

Solution:

 1 class Element implements Comparable<Element>{
 2     int val;
 3     int index;
 4     public Element(int v, int i){
 5         val = v;
 6         index = i;
 7     }
 8 
 9     public int compareTo(Element other){
10         return this.val - other.val;
11     }
12 
13     public int getIndex(){
14         return index;
15     }
16 
17     public int getValue(){
18         return val;
19     }
20 }
21 
22 public class Solution {
23     /**
24      * @param nums: A list of integers
25      * @return: A list of integers includes the index of the first number
26      *          and the index of the last number
27      */
28     public ArrayList<Integer> subarraySumClosest(int[] nums) {
29         ArrayList<Integer> res = new ArrayList<Integer>();
30         if (nums.length==0) return res;
31 
32         Element[] sums = new Element[nums.length+1];
33         sums[0] = new Element(0,-1);
34         int sum = 0;
35         for (int i=0;i<nums.length;i++){
36             sum += nums[i];
37             sums[i+1] = new Element(sum,i);
38         }
39 
40         Arrays.sort(sums);
41         int min = Math.abs(sums[0].getValue() - sums[1].getValue());
42         int start =  Math.min(sums[0].getIndex(), sums[1].getIndex())+1;
43         int end = Math.max(sums[0].getIndex(), sums[1].getIndex());
44         for (int i=1;i<nums.length;i++){
45             int diff = Math.abs(sums[i].getValue() - sums[i+1].getValue());
46             if (diff<min){
47                 min = diff;
48                 start = Math.min(sums[i].getIndex(), sums[i+1].getIndex())+1;
49                 end  = Math.max(sums[i].getIndex(), sums[i+1].getIndex());
50             }
51         }
52 
53         res.add(start);
54         res.add(end);
55         return res;
56     }
57 }