Question

这个请求不使用二维数组。考虑以下结构的一维数组，其中索引从零开始，长度从一开始：

第 0 部分是单个单元格的数组。

第 1 节

子阵列 1
- 索引0
- 索引1
- ...
- 索引n - 1
子阵列 2
- 索引0
- 索引1
- ...
- 索引n - 1
...
子阵列k
- 索引0
- 索引1
- ...
- 索引m - 1

第 2 节和第 3 节的结构与第 1 节相同。

有多个阵列拼凑在一起以模拟更大的阵列。总共有四个部分：部分 0 将始终具有单个单元格，部分 1、2 和 3 将每个具有 k 子数组。子数组 1 到 k-1（最后一个子数组除外）的长度为 n。最后一个子数组将包含 m 元素和 m < n。

目标：给定一个从零开始的累积索引input，使用索引公式找到一个section的子数组中对应的索引，最少的if语句和没有循环。

这是一个例子：

让 k = 3、n = 8 和 m = 5 都从一开始。以下是第 0 节和第 1 节的正确输出：

 input =  0 → section = 0, array = 0, subIndex = 0

 input =  1 → section = 1, array = 0, subIndex = 0
 input =  2 → section = 1, array = 0, subIndex = 1
 input =  3 → section = 1, array = 0, subIndex = 2
 input =  4 → section = 1, array = 0, subIndex = 3
 input =  5 → section = 1, array = 0, subIndex = 4
 input =  6 → section = 1, array = 0, subIndex = 5
 input =  7 → section = 1, array = 0, subIndex = 6
 input =  8 → section = 1, array = 0, subIndex = 7
 input =  9 → section = 1, array = 1, subIndex = 0
 input = 10 → section = 1, array = 1, subIndex = 1
 input = 11 → section = 1, array = 1, subIndex = 2
 input = 12 → section = 1, array = 1, subIndex = 3
 input = 13 → section = 1, array = 1, subIndex = 4
 input = 14 → section = 1, array = 1, subIndex = 5
 input = 15 → section = 1, array = 1, subIndex = 6
 input = 16 → section = 1, array = 1, subIndex = 7
 input = 17 → section = 1, array = 2, subIndex = 0
 input = 18 → section = 1, array = 2, subIndex = 1
 input = 19 → section = 1, array = 2, subIndex = 2
 input = 20 → section = 1, array = 2, subIndex = 3
 input = 21 → section = 1, array = 2, subIndex = 4

接下来的条目应该是：

input = 22 → section = 2, array = 0, subIndex = 0
...
input = 30 → section = 2, array = 1, subIndex = 0
...
input = 38 → section = 2, array = 2, subIndex = 0
...
input = 42 → section = 2, array = 2, subIndex = 4

input = 43 → section = 3, array = 0, subIndex = 0
...
input = 51 → section = 3, array = 1, subIndex = 0
...
input = 59 → section = 3, array = 2, subIndex = 0
...
input = 63 → section = 3, array = 2, subIndex = 4

截至目前，我的带有单个 if 语句的公式只能获取第一部分（不是第 2 部分或第 3 部分）。希望找到一组公式来实现目标，而无需使用 if 语句或循环。这是我目前所拥有的：

public class Test {

    public Test() {
        for (int i = 0; i < 63; i++) {
            getIndex(i);
        }
    }

    public void getIndex(int i) {
        if (i == 0) {//Aligns output. I couldn't factor it in the formula
            System.out.println("input = " + i + " → " +
                    "section = " + 0 +
                    ", array = " + 0 +
                    ", subIndex = " + 0);
            return;
        }
        int k             = 3;//Number of subarrays for each section
        int n             = 8;//all subarray sizes except last one
        int m             = 5;//last subarray size
        int sectionLength = (k - 1) * n + m;//number of cells in a single
                                            //section (other than section 0).

        //obtaining the values:
        int section  = (i - 1) / sectionLength + 1;//the section (1, 2 or 3)
        int subarray = (i - 1) / (section * n);
        int index    = ((i - 1) % n);
        System.out.println("input = " + i + " → " +
                "section = " + section +
                ", array = " + subarray +
                ", subIndex = " + index);
    }

    public static void main(String[] args) {
        Test t = new Test();
    }
}

Answer 1

public class Test {
    public static void main(String[] args) {
        
        // all your constants
        final int TOTAL_SECTION = 63;
        final int SUBARRAY_PER_SECTION = 3;
        final int SUBARRAY_SIZE = 8;
        final int FINAL_SUBARRAY_SIZE = 5;

        // keep track of current input number, increase one after each input
        int inputNumber = 0;

        for (int i = 0; i < TOTAL_SECTION; i++) {
            
            // if it is 0, all the value is 0, so just print the defualt format with all 0
            if (i == 0) {
                print(inputNumber++, 0 , 0, 0);
                System.out.println();
                continue;
            }

            for (int j = 0; j < SUBARRAY_PER_SECTION; j++) {
                
                // if it is last subarray, use the final subarray size
                if (j == SUBARRAY_PER_SECTION - 1) {
                    for (int k = 0; k < FINAL_SUBARRAY_SIZE; k++) {
                        print(inputNumber++, i, j, k);
                    }
                    break;
                }

                // otherwise, just use the normal subarray size
                for (int k = 0; k < SUBARRAY_SIZE; k++) {
                    print(inputNumber++, i, j, k);
                }
            }

            // print a new line after each section done
            System.out.println();
        }
    }

    // format of the message, use this method and give the 4 values needed as parameters
    private static void print(int inputNumber, int sectionNumber, int arrayNumber, int subIndex) {
        System.out.printf("input = %2d → section = %d, array = %d, subIndex = %d %n", inputNumber, sectionNumber, arrayNumber, subIndex);
    }
}

查找由多个组合子数组组成的数组的索引

1 个答案: