Java 计算数组中不是每个元素的除数的元素数

我试图从可理解性的角度来理解问题的解决方案。这个问题要求计算数组中不是每个元素的除数的元素数。下文提供了完整的说明

You are given a non-empty zero-indexed array A consisting of N integers.
    For each number A[i] such that 0 ≤ i < N, we want to count the number of elements of the array that are not the divisors of A[i]. We say that these elements are non-divisors.
    For example, consider integer N = 5 and array A such that:
        A[0] = 3
        A[1] = 1
        A[2] = 2
        A[3] = 3
        A[4] = 6
    For the following elements:
    A[0] = 3, the non-divisors are: 2, 6,
    A[1] = 1, the non-divisors are: 3, 2, 3, 6,
    A[2] = 2, the non-divisors are: 3, 3, 6,
    A[3] = 3, the non-divisors are: 2, 6,
    A[6] = 6, there aren't any non-divisors.
    Write a function:
    class Solution { public int[] solution(int[] A); }
    that, given a non-empty zero-indexed array A consisting of N integers, returns a sequence of integers representing the amount of non-divisors.
    The sequence should be returned as:
    a structure Results (in C), or
    a vector of integers (in C++), or
    a record Results (in Pascal), or
    an array of integers (in any other programming language).
    For example, given:
        A[0] = 3
        A[1] = 1
        A[2] = 2
        A[3] = 3
        A[4] = 6
    the function should return [2, 4, 3, 2, 0], as explained above.
    Assume that:
    N is an integer within the range [1..50,000];
    each element of array A is an integer within the range [1..2 * N].
    Complexity:
    expected worst-case time complexity is O(N*log(N));
    expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
    Elements of input arrays can be modified.

将为您提供一个由N个整数组成的非空零索引数组。
对于每个数字A[i]，0≤ i

我也有一个解决办法

// int[] A = {3, 1, 2, 3, 6};
public static int[] solution(int[] A) {

    int[][] D = new int[2 * A.length + 1][2];
    int[] res = new int[A.length];      

        //----- 
        // 0 1 
        // 0 0 
        // 1 -1 
        // 1 -1 
        // 2 -1 
        // 0 0 
        // 0 0 
        // 1 -1 
        // 0 0 
        // 0 0 
        // 0 0 
        // 0 0
        //-----

    for (int i = 0; i < A.length; i++) {
        // D[A[i]][0]++;

        D[A[i]][0] = D[A[i]][0] + 1;
        D[A[i]][1] = -1;
    }


    for (int i = 0; i < A.length; i++){

        if(D[A[i]][1]==-1){

            D[A[i]][1]=0;

            for (int j = 1; j*j <= A[i]; j++) {

                if(A[i] % j == 0) {

                    // D[A[i]][1] = D[A[i]][1] + D[j][0];
                    D[A[i]][1] += D[j][0];

                    if (A[i]/j != j){
                        D[A[i]][1]+= D[A[i]/j][0];
                    }
                }                   
            }
        }
    }

    for (int i = 0; i < A.length; i++) {
        res[i] = A.length - D[A[i]][1]; 
    }

    return res;
}   

//int[]A={3,1,2,3,6}；
公共静态int[]解决方案（int[]A）{
int[][]D=新int[2*A.length+1][2]；
int[]res=新的int[A.长度]；
//----- 
// 0 1 
// 0 0 
// 1 -1 
// 1 -1 
// 2 -1 
// 0 0 
// 0 0 
// 1 -1 
// 0 0 
// 0 0 
// 0 0 
// 0 0
//-----
for（int i=0；i对于（int j=1；j*j，
D
数据结构/矩阵是这样的：
0th
列和
jth
行计算
j
在数组
A
中出现的次数。换句话说，
D[A[j][0]
是
A[j]
值在数组中的次数

循环之后，
1st
列和
kth
行计算数组中划分
A[k]
的元素数。换句话说，
D[A[k][1]
是数组中
A[k]
的除数

最后的结果是，
r[j]
只是
r[j]=（A.length）-D[A[j][1]
。因为我们想要的是不是除数的元素数

为什么循环工作

如果
A[i]%j==0
，那么我们要做的是计算
j
出现在
A
中的次数，然后将其添加到
D[A[i][1]
。这就是为什么你有行
D[A[i][1]+=D[j][0]；
。此外
A[i]/j
也将是一个不同的因素（除非
A[i]=j
）

数学部分来证明集合{A，B|A*B=N&AThe
j*j这非常有用。非常感谢你。
            for (int j = 1; j*j <= A[i]; j++) {

                if(A[i] % j == 0) {

                    // D[A[i]][1] = D[A[i]][1] + D[j][0];
                    D[A[i]][1] += D[j][0];

                    if (A[i]/j != j){
                        D[A[i]][1]+= D[A[i]/j][0];
                    }
                }                   
            }