java - 其值是每个其他整数的乘积的数组

Question

这个问题已经有答案了:

Given an array of numbers, return array of products of all other numbers (no division) (48 个回答)

已关闭 6 年前。

我在电话中被问到以下面试问题:

Given an array of integers, produce an array whose values are the product of every other integer 

excluding the current index. 

Example: 

[4, 3, 2, 8] -> [3*2*8, 4*2*8, 4*3*8, 4*3*2] -> [48, 64, 96, 24]

我给出了以下答案

import java.math.BigInteger;
import java.util.Arrays;

public class ProductOfAnArray {

    public static void main(String[] args) {

        try {
            System.out.println(Arrays.toString(ProductOfAnArray
                    .calcArray(new int[] { 4, 3, 2, 8 })));
            System.out.println(Arrays.toString(ProductOfAnArray
                    .calcArray(new int[] { 4, 0, 2, 8 })));
            System.out.println(Arrays.toString(ProductOfAnArray
                    .calcArray(new int[] { 4, 0, 2, 0 })));
            System.out.println(Arrays.toString(ProductOfAnArray
                    .calcArray(new int[] {})));
            System.out
                    .println(Arrays.toString(ProductOfAnArray
                            .calcArray(new int[] { 4, 3, 2, 8, 3, 2, 4, 6, 7,
                                    3, 2, 4 })));
            System.out
                    .println(Arrays.toString(ProductOfAnArray
                            .calcArray(new int[] { 4, 3, 2, 8, 3, 2, 4, 6, 7,
                                    3, 2, 4 })));
            System.out.println(Arrays.toString(ProductOfAnArray
                    .calcArray(new int[] { 4432432, 23423423, 34234, 23423428,
                            4324243, 24232, 2342344, 64234234, 4324247,
                            4234233, 234422, 234244 })));
        } catch (Exception e) {
            // debug exception here and log.
        }
    }

    /*
     * Problem: Given an array of integers, produce an array whose values are
     * the product of every other integer excluding the current index.
     * 
     * Assumptions. Input array cannot be modified. input is an integer array
     * "produce an array" - type not specified for output array
     * 
     * Logic explanation:
     * 
     * Assume we have array [a,b,c,d] Let multiple be multiple of each element
     * in array. Hence multiple = 0 if one of the element is 0; To produce the
     * output. Ans at i -> multiple divided by the value at i. if 2 numbers are
     * 0 then entire output will be 0 because atleast one 0 will be a multiple
     * if 1 number is 0 then every thing else will be 0 except that index whole
     * value is to be determined
     * 
     */
    public static BigInteger[] calcArray(final int[] inp) throws Exception {
        if (inp == null)
            throw new Exception("input is null");

        int cnt = 0;
        BigInteger multiple = new BigInteger("1");
        boolean foundZero = false;

        for (int i : inp) {
            if (i == 0) {
                cnt++;
                foundZero = true;
                if (cnt < 2)
                    continue;
                else
                    break;
            }
            multiple = multiple.multiply(BigInteger.valueOf(i));
        }

        BigInteger ans[] = new BigInteger[inp.length];

        for (int i = 0; i < inp.length; i++) {
            if (foundZero) {
                if (cnt < 2) {
                    ans[i] = (inp[i] == 0) ? multiple : new BigInteger("0");
                } else {
                    ans[i] = new BigInteger("0");
                }
            } else {
                ans[i] = multiple.divide(BigInteger.valueOf(inp[i]));
            }
        }
        return ans;
    }

}

但是我没有被选中。我想获得对我的答案的反馈并看看有什么问题。

谢谢。

Answer 1

我过去也尝试过同样的方法。没有选择:)。我试图在第一个循环中捕获零索引(如果有的话)。并简单地将乘积分配给该索引(我使用双数组，默认值为 0) - 因此如果找到 0，则无需再次迭代。

并在第一个循环中检查乘积是否为无穷大Double.isInfinite，如果是，则中断循环 - 没有必要找到剩余的乘积(假设输入是大量的大数字)

public static double[] arrayProducts(int[] input) {
    int length = input.length;
    double product = 1;
    boolean zeroFound = false;
    int zeroIndex = 0;
    double[] output = null;
    for (int i = 0; i < length; i++) {
        if (input[i] == 0) {
            if (zeroFound) {
                throw new ProductArrayException(0, true, zeroIndex,
                        ZERO_FOUND_EXPECTION);
            }
            zeroFound = true;
            zeroIndex = i;
        } else {
            product = product * input[i];
            if (Double.isInfinite(product)) {
                throw new ProductArrayException(0, true, zeroIndex,
                        INFINITY_FOUND_EXPECTION);
            }
        }
    }

    if (zeroFound) {
        throw new ProductArrayException(product, false, zeroIndex,
                ZERO_FOUND_EXPECTION);
    } else {
        output = new double[length];
        for (int i = 0; i < length; i++) {
            output[i] = product / input[i];
        }
    }
    return output;
}