java解决小和问题

第一种暴力解决 O(N^2)

import java.util.Scanner;

public class code10_SmallSum {
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine().toString();
        String[] arr = str.split(" ");
        int[] b=new int[arr.length];
        for(int i=0;i 
第二种用归并排序的思想解决O(N*logN)
 

import java.util.Scanner;

public class code11_smallSum2 {

    public static int smallSum(int[] arr) {
        if (arr == null || arr.length < 2) {
            return 0;
        }
        return process(arr, 0, arr.length - 1);
    }
    public static int process(int[] arr, int l, int r) {
        if (l == r) {
            return 0;
        }
        // l < r
        int mid = l + ((r - l) >> 1);
        return
                process(arr, l, mid)
                        +
                        process(arr, mid + 1, r)
                        +
                        merge(arr, l, mid, r);
    }
    public static int merge(int[] arr, int L, int m, int r) {
        int[] help = new int[r - L + 1];
        int i = 0;
        int p1 = L;
        int p2 = m + 1;
        int res = 0;
        while (p1 <= m && p2 <= r) {
            //这个主要是先取右边的
            res += arr[p1] < arr[p2] ? (r - p2 + 1) * arr[p1] : 0;
            help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
        }
        while (p1 <= m) {
            help[i++] = arr[p1++];
        }
        while (p2 <= r) {
            help[i++] = arr[p2++];
        }
        for (i = 0; i < help.length; i++) {
            arr[L + i] = help[i];
        }
        return res;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine().toString();
        String[] arr = str.split(" ");
        int[] b=new int[arr.length];
        for(int i=0;i 
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