给定一个由N个整数组成的数组arr [] ，任务是构造一个大小相同的乘积数组，而不使用除法运算符( ‘/’ )，以使每个数组元素等于arr []的所有元素的乘积除了arr [i] 。
例子：
Input: arr[] = {10, 3, 5, 6, 2}
Output: 180 600 360 300 900
Explanation:
3 * 5 * 6 * 2 is the product of all array elements except 10 is 180
10 * 5 * 6 * 2 is the product of all array elements except 3 is 600.
10 * 3 * 6 * 2 is the product of all array elements except 5 is 360.
10 * 3 * 5 * 2 is the product of all array elements except 6 is 300.
10 * 3 * 6 * 5 is the product of all array elements except 2 is 9.
Input: arr[] = {1, 2, 1, 3, 4}
Output: 24 12 24 8 6
方法：想法是使用log()和exp()函数代替log10()和pow()。以下是关于同一方面的一些观察结果：
- 假设M是所有数组元素的乘积，则在第i个位置的输出数组元素将等于M / arr [i]。
- 可以通过使用对数和exp函数的属性来执行两个数的除法。
- 没有为小于零的数字定义对数函数，因此要分别维护这种情况。
请按照以下步骤解决问题：
- 初始化两个变量，例如product = 1和Z = 1 ，以存储数组和零元素计数的乘积。
- 如果arr [i]不等于0，则遍历数组并将乘积乘以arr [i] 。否则，将Z的计数增加1 。
- 遍历数组arr []并执行以下操作：
- 如果Z为1且arr [i]不为零，则将arr [i]更新为arr [i] = 0并 继续。
- 否则，如果Z为1且arr [i]为0，则将arr [i]更新为乘积并继续。
- 否则，如果Z大于1，则将arr [i]分配为0并继续。
- 现在，使用上述公式找到abs(product)/ abs(arr [i])的值，并将其存储在变量curr中。
- 如果arr [i]和product的值为负，或者arr [i]和product为正，则将arr [i]分配为curr 。
- 否则，将arr [i]分配为-1 * curr 。
- 完成上述步骤后，打印数组arr [] 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Fucntion to form product array
// with O(n) time and O(1) space
void productExceptSelf(int arr[],
int N)
{
// Stores the product of array
int product = 1;
// Stores the count of zeros
int z = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// If arr[i] is not zero
if (arr[i])
product *= arr[i];
// If arr[i] is zero then
// increment count of z by 1
z += (arr[i] == 0);
}
// Stores the absolute value
// of the product
int a = abs(product), b;
for (int i = 0; i < N; i++) {
// If Z is equal to 1
if (z == 1) {
// If arr[i] is not zero
if (arr[i])
arr[i] = 0;
// Else
else
arr[i] = product;
continue;
}
// If count of 0s at least 2
else if (z > 1) {
// Assign arr[i] = 0
arr[i] = 0;
continue;
}
// Store absoulute value of arr[i]
int b = abs(arr[i]);
// Find the value of a/b
int curr = round(exp(log(a) - log(b)));
// If arr[i] and product both
// are less than zero
if (arr[i] < 0 && product < 0)
arr[i] = curr;
// If arr[i] and product both
// are greater than zero
else if (arr[i] > 0 && product > 0)
arr[i] = curr;
// Else
else
arr[i] = -1 * curr;
}
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
cout << arr[i] << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 10, 3, 5, 6, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
productExceptSelf(arr, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Fucntion to form product array
// with O(n) time and O(1) space
static void productExceptSelf(int arr[],
int N)
{
// Stores the product of array
int product = 1;
// Stores the count of zeros
int z = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// If arr[i] is not zero
if (arr[i] != 0)
product *= arr[i];
// If arr[i] is zero then
// increment count of z by 1
if (arr[i] == 0)
z += 1;
}
// Stores the absolute value
// of the product
int a = Math.abs(product);
for (int i = 0; i < N; i++) {
// If Z is equal to 1
if (z == 1) {
// If arr[i] is not zero
if (arr[i] != 0)
arr[i] = 0;
// Else
else
arr[i] = product;
continue;
}
// If count of 0s at least 2
else if (z > 1) {
// Assign arr[i] = 0
arr[i] = 0;
continue;
}
// Store absoulute value of arr[i]
int b = Math.abs(arr[i]);
// Find the value of a/b
int curr = (int)Math.round(Math.exp(Math.log(a) - Math.log(b)));
// If arr[i] and product both
// are less than zero
if (arr[i] < 0 && product < 0)
arr[i] = curr;
// If arr[i] and product both
// are greater than zero
else if (arr[i] > 0 && product > 0)
arr[i] = curr;
// Else
else
arr[i] = -1 * curr;
}
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
System.out.print(arr[i] + " ");
}
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 10, 3, 5, 6, 2 };
int N = arr.length;
// Function Call
productExceptSelf(arr, N);
}
}
// This code is contributed by splevel62.Python3
# Python 3 program for the above approach
import math
# Fucntion to form product array
# with O(n) time and O(1) space
def productExceptSelf(arr, N) :
# Stores the product of array
product = 1
# Stores the count of zeros
z = 0
# Traverse the array
for i in range(N):
# If arr[i] is not zero
if (arr[i] != 0) :
product *= arr[i]
# If arr[i] is zero then
# increment count of z by 1
if(arr[i] == 0):
z += 1
# Stores the absolute value
# of the product
a = abs(product)
for i in range(N):
# If Z is equal to 1
if (z == 1) :
# If arr[i] is not zero
if (arr[i] != 0) :
arr[i] = 0
# Else
else :
arr[i] = product
continue
# If count of 0s at least 2
elif (z > 1) :
# Assign arr[i] = 0
arr[i] = 0
continue
# Store absoulute value of arr[i]
b = abs(arr[i])
# Find the value of a/b
curr = round(math.exp(math.log(a) - math.log(b)))
# If arr[i] and product both
# are less than zero
if (arr[i] < 0 and product < 0):
arr[i] = curr
# If arr[i] and product both
# are greater than zero
elif (arr[i] > 0 and product > 0):
arr[i] = curr
# Else
else:
arr[i] = -1 * curr
# Traverse the array arr[]
for i in range(N):
print(arr[i], end = " ")
# Driver Code
arr = [ 10, 3, 5, 6, 2 ]
N = len(arr)
# Function Call
productExceptSelf(arr, N)
# This code is contributed by code_hunt.C#
// C# program for the above approach
using System;
class GFG
{
// Fucntion to form product array
// with O(n) time and O(1) space
static void productExceptSelf(int[] arr,
int N)
{
// Stores the product of array
int product = 1;
// Stores the count of zeros
int z = 0;
// Traverse the array
for (int i = 0; i < N; i++)
{
// If arr[i] is not zero
if (arr[i] != 0)
product *= arr[i];
// If arr[i] is zero then
// increment count of z by 1
if (arr[i] == 0)
z += 1;
}
// Stores the absolute value
// of the product
int a = Math.Abs(product);
for (int i = 0; i < N; i++)
{
// If Z is equal to 1
if (z == 1)
{
// If arr[i] is not zero
if (arr[i] != 0)
arr[i] = 0;
// Else
else
arr[i] = product;
continue;
}
// If count of 0s at least 2
else if (z > 1)
{
// Assign arr[i] = 0
arr[i] = 0;
continue;
}
// Store absoulute value of arr[i]
int b = Math.Abs(arr[i]);
// Find the value of a/b
int curr = (int)Math.Round(Math.Exp(Math.Log(a) - Math.Log(b)));
// If arr[i] and product both
// are less than zero
if (arr[i] < 0 && product < 0)
arr[i] = curr;
// If arr[i] and product both
// are greater than zero
else if (arr[i] > 0 && product > 0)
arr[i] = curr;
// Else
else
arr[i] = -1 * curr;
}
// Traverse the array arr[]
for (int i = 0; i < N; i++)
{
Console.Write(arr[i] + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 10, 3, 5, 6, 2 };
int N = arr.Length;
// Function Call
productExceptSelf(arr, N);
}
}
// This code is contributed by sanjoy_62.输出：
180 600 360 300 900时间复杂度： O(N)
辅助空间： O(1)
‘
替代方法：有关替代方法，请参阅本文的先前文章：
- 产品阵列之谜
- 产品阵列难题|套装2