通过电话我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;
}
}
但我没被选中。我希望得到我的答案的反馈，看看有什么不对。
谢谢。
答案 0 :(得分：1)
在第一个循环中检查产品是无穷大Double.isInfinite，如果是，则打破循环 - 没有找到剩余产品的点（假设输入是高数量的大数量）
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;
}
答案 1 :(得分：0)
这可能是面试官追求的解决方案：
int[] arr = new int[]{4,3,2,8};
int[] newarr = new int[arr.length];
int product = 1;
for (int i=0; i < arr.length; i++) {
product = product * arr[i];
}
for (int i=0; i < arr.length; i++) {
newarr[i] = product / arr[i];
System.out.print(newarr[i] + " ");
}
答案 2 :(得分：0)
给定整数数组arr[]，
int[] arr = { 4, 3, 2, 8 };
int[] prod = new int[arr.length];
for (int i = 0; i < arr.length; i++)
{
prod[i] = 1;
for (int j = 0; j < arr.length; j++)
if (j != i) // Exclude element at the current index
prod[i] *= arr[j];
}
生成的产品数组为prod[]。
答案 3 :(得分：0)
通常对于这个问题，不计算除法的答案，它必须是一个有效的时间和空间算法。
看下面产生O（n）时间＆amp; O（n）空间
public int[] otherIndexProduct(int array[]) {
int results[] = new int[array.length];
results[0] = 1;
for (int i = 0; i < array.length - 1; i++) {
results[i + 1] = results[i] * array[i];
}
int before = 1;
for (int i = array.length - 1; i >= 0; i--) {
results[i] *= before;
before *= array[i];
}
return results;
}
有一种有趣的另一种方法来实现递归，它也可以在不使用其他数组的情况下更新输入数组
public int product(int array[], int post, int index) {
int prev = 1;
if (index < array.length) {
prev = product(array, post * array[index], index + 1);
int current = array[index];
array[index] = post * prev;
prev *= current;
}
return prev;
}