java - 数组，其值是每隔一个整数的乘积

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

（45个答案）


3年前关闭。





通过电话向我询问以下面试问题：

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。我想获得我的答案的反馈，看看有什么问题。

谢谢。

最佳答案

我过去也尝试过相同的方法。没有选择:)。
我试图在第一循环中捕获零索引。并简单地将乘积分配给该索引（我使用了双精度数组，默认值为0）-因此，如果找到一个零，则无需再次进行迭代。

并在第一个循环中检查乘积是否为无穷大Double.isInfinite，如果是，则打破循环-找不到剩余的乘积点（假定输入是大容量的大数字）

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;
}