The longest consecutive subarray sum is 0

topic

Find the longest contiguous subarray whose sum is 0.

analyze

Exhaustive method: scan the array twice to find the length of all subarrays whose sum is 0, and then select the largest one. time complexity$O(n^2)$。

Improvement: Given an array such as 1, 2, 3, 4, -1, -2, -4, -3, 1, 1,1,1,1, find its accumulated sum array 1,3,6, 10,9,7,3,0,1,2,3,4,5. Sequentially scan the array and record its index value index when 0 is encountered, then the length is index-0+1. If it encounters a non-zero value, then scan at the left end again to find a number that is equal to it, then the length between these two numbers is the length of the other satisfying condition. Return the longest one. time complexity$O(n^2)$, a hash table can be used to make the time complexity O(n).

code

import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        //int[] a = new int[]{1, 2, 3, 4, -1, -2, -4, -3, 1, 2};
        //int[] a = new int[]{1, 2, 3, 4, -1, -2, -4, -3, 1, 1};
        Scanner cin = new Scanner(System.in);
        String s = cin.nextLine().trim();
        String[] numStr = s.split(" ");
        int[] num = new int[numStr.length];
        for(int i = 0;i < numStr.length;i++){
            num[i] = Integer.parseInt(numStr[i]);
        }

        int[] res = maxSubZero(num);
        for(int var: res)
            System.out.println(var);

        cin.close();
    }

    public static int [] maxSubZero(int[] a){
        int[] sum = new int[a.length];
        int gstart = 0, gend = 0;

        sum[0] = a[0];
        for(int i = 1; i < a.length; i++){
            sum[i] = sum[i-1] + a[i];
        }

        for(int i = 0; i < sum.length; i++){
            int start = 0, end = 0;
            for(int j = 0; j < i; j++){
                if(sum[i] == 0){
                    end = i;
                }
                else if(sum[i] == sum[j]){
                    start = j+1;
                    end = i;
                    //System.out.println(start + " " + end);
                }
                if(gend - gstart < end - start){
                    gend = end;
                    gstart = start;
                }
            }
        }

        int[] res = new int[gend-gstart+1];
        for(int k = gstart; k <= gend; k++){
            res[k-gstart] = a[k];
        }
        return res;
    }
}