本文概述
给定一个包含n个整数的数组arr[],构造一个乘积数组prod[](大小相同),使prod[i]等于arr[]中除arr[i]之外的所有元素的乘积。在O(n)范围内,不需要除法运算。
例子:
Input: arr[] = {10, 3, 5, 6, 2}
Output: prod[] = {180, 600, 360, 300, 900}
The elements of output array are
{3*5*6*2, 10*5*6*2, 10*3*6*2, 10*3*5*2, 10*3*5*6}
Input: arr[] = {1, 2, 1, 3, 4}
Output: prod[] = {24, 12, 24, 8, 6}
The elements of output array are
{3*4*1*2, 1*1*3*4, 4*3*2*1, 1*1*4*2, 1*1*3*2}
在乘积数组问题|S1中已经有一个讨论过的O(n)方法。前面的方法使用额外的O(n)空间来构造product数组。
解决方案1:使用log属性。
方法:在这篇文章中, 已经讨论了一种更好的方法, 该方法使用log属性查找除特定索引处的数组所有元素的乘积。这种方法不占用额外的空间。
使用log的属性将大数相乘
x = a * b * c * d
log(x) = log(a * b * c * d)
log(x) = log(a) + log(b) + log(c) + log(d)
x = antilog(log(a) + log(b) + log(c) + log(d))
所以这个想法很简单,
遍历数组并找到所有元素的对数和,
log(a[0]) + log(a[1]) +
.. + log(a[n-1])
然后再次遍历数组, 并使用该公式查找乘积。
antilog((log(a[0]) + log(a[1]) +
.. + log(a[n-1])) - log(a[i]))
这等于除a [i]以外的所有元素的乘积, 即antilog(sumlog(a [i]))。
C ++
// C++ program for product array puzzle
// with O(n) time and O(1) space.
#include <bits/stdc++.h>
using namespace std;
// epsilon value to maintain precision
#define EPS 1e-9
void productPuzzle( int a[], int n)
{
// to hold sum of all values
long double sum = 0;
for ( int i = 0; i < n; i++)
sum += ( long double ) log10 (a[i]);
// output product for each index
// antilog to find original product value
for ( int i = 0; i < n; i++)
cout << ( int )(EPS + pow (( long double )10.00, sum - log10 (a[i])))
<< " " ;
}
// Driver code
int main()
{
int a[] = { 10, 3, 5, 6, 2 };
int n = sizeof (a) / sizeof (a[0]);
cout << "The product array is: \n" ;
productPuzzle(a, n);
return 0;
}
Java
// Java program for product array puzzle
// with O(n) time and O(1) space.
public class Array_puzzle_2 {
// epsilon value to maintain precision
static final double EPS = 1e- 9 ;
static void productPuzzle( int a[], int n)
{
// to hold sum of all values
double sum = 0 ;
for ( int i = 0 ; i < n; i++)
sum += Math.log10(a[i]);
// output product for each index
// anti log to find original product value
for ( int i = 0 ; i < n; i++)
System.out.print(
( int )(EPS
+ Math.pow(
10.00 , sum
- Math.log10(a[i])))
+ " " );
}
// Driver code
public static void main(String args[])
{
int a[] = { 10 , 3 , 5 , 6 , 2 };
int n = a.length;
System.out.println( "The product array is: " );
productPuzzle(a, n);
}
}
// This code is contributed by Sumit Ghosh
python
# Python program for product array puzzle
# with O(n) time and O(1) space.
import math
# epsilon value to maintain precision
EPS = 1e - 9
def productPuzzle(a, n):
# to hold sum of all values
sum = 0
for i in range (n):
sum + = math.log10(a[i])
# output product for each index
# antilog to find original product value
for i in range (n):
print int ((EPS + pow ( 10.00 , sum - math.log10(a[i])))), return
# Driver code
a = [ 10 , 3 , 5 , 6 , 2 ]
n = len (a)
print "The product array is: "
productPuzzle(a, n)
# This code is contributed by Sachin Bisht
C#
// C# program for product
// array puzzle with O(n)
// time and O(1) space.
using System;
class GFG {
// epsilon value to
// maintain precision
static double EPS = 1e-9;
static void productPuzzle( int [] a, int n)
{
// to hold sum of all values
double sum = 0;
for ( int i = 0; i < n; i++)
sum += Math.Log10(a[i]);
// output product for each
// index anti log to find
// original product value
for ( int i = 0; i < n; i++)
Console.Write(( int )(EPS + Math.Pow(10.00, sum - Math.Log10(a[i]))) + " " );
}
// Driver code
public static void Main()
{
int [] a = { 10, 3, 5, 6, 2 };
int n = a.Length;
Console.WriteLine( "The product array is: " );
productPuzzle(a, n);
}
}
// This code is contributed by mits
的PHP
<?php
// PHP program for product array puzzle
// with O(n) time and O(1) space.
// epsilon value to maintain precision
$EPS =1e-9;
function productPuzzle( $a , $n )
{
global $EPS ;
// to hold sum of all values
$sum = 0;
for ( $i = 0; $i < $n ; $i ++)
$sum += (double)log10( $a [ $i ]);
// output product for each index
// antilog to find original product value
for ( $i = 0; $i < $n ; $i ++)
echo (int)( $EPS + pow((double)10.00, $sum - log10( $a [ $i ]))). " " ;
}
// Driver code
$a = array (10, 3, 5, 6, 2 );
$n = count ( $a );
echo "The product array is: \n" ;
productPuzzle( $a , $n );
// This code is contributed by mits
?>
输出如下:
The product array is:
180 600 360 300 900
复杂度分析:
- 时间复杂度:O(n)。
仅需要两次遍历数组。 - 空间复杂度:O(1)。
不需要额外的空间。
该方法由Abhishek Rajput提供。
替代方法:这是通过使用pow()函数解决上述问题的另一种方法, 该方法不使用除法并且可以在O(n)时间内工作。
遍历数组并找到数组中所有元素的乘积。将乘积存储在变量中。
然后再次遍历数组, 并使用公式(product * pow(a [i], -1))查找除该数字以外的所有元素的乘积
C ++
// C++ program for product array puzzle
// with O(n) time and O(1) space.
#include <bits/stdc++.h>
using namespace std;
// Solve function which prints the answer
void solve( int arr[], int n)
{
// Initialize a variable to store the
// total product of the array elements
int prod = 1;
for ( int i = 0; i < n; i++)
prod *= arr[i];
// we know x/y mathematically is same
// as x*(y to power -1)
for ( int i = 0; i < n; i++)
cout << prod * ( int ) pow (
arr[i], -1)
<< " " ;
}
// Driver Code
int main()
{
int arr[] = { 10, 3, 5, 6, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
solve(arr, n);
return 0;
}
// This code is contributed by Sitesh Roy
Java
// Java program for product array puzzle
// with O(n) time and O(1) space.
public class ArrayPuzzle {
static void solve( int arr[], int n)
{
// Initialize a variable to store the
// total product of the array elements
int prod = 1 ;
for ( int i = 0 ; i < n; i++)
prod *= arr[i];
// we know x/y mathematically is same
// as x*(y to power -1)
for ( int i = 0 ; i < n; i++)
System.out.print(
( int )prod * Math.pow(arr[i], - 1 ) + " " );
}
// Driver code
public static void main(String args[])
{
int arr[] = { 10 , 3 , 5 , 6 , 2 };
int n = arr.length;
solve(arr, n);
}
}
// This code is contributed by Sitesh Roy
Python3
# Python program for product array puzzle
# with O(n) time and O(1) space.
def solve(arr, n):
# Initialize a variable to store the
# total product of the array elements
prod = 1
for i in arr:
prod * = i
# we know x / y mathematically is same
# as x*(y to power -1)
for i in arr:
print ( int (prod * (i * * - 1 )), end = " " )
# Driver Code
arr = [ 10 , 3 , 5 , 6 , 2 ]
n = len (arr)
solve(arr, n)
# This code is contributed by Sitesh Roy
C#
// C# program for product array puzzle
// with O(n) time and O(1) space.
using System;
class GFG {
public
class ArrayPuzzle {
static void solve( int [] arr, int n)
{
// Initialize a variable to store the
// total product of the array elements
int prod = 1;
for ( int i = 0; i < n; i++)
prod *= arr[i];
// we know x/y mathematically is same
// as x*(y to power -1)
for ( int i = 0; i < n; i++)
Console.Write(
( int )prod * Math.Pow(arr[i], -1) + " " );
}
// Driver code
static public void Main()
{
int [] arr = { 10, 3, 5, 6, 2 };
int n = arr.Length;
solve(arr, n);
}
}
}
// This code is contributed by shivanisinghss2110
输出如下:
180 600 360 300 900
复杂度分析:
- 时间复杂度:O(n)。
仅需要两次遍历数组。 - 空间复杂度:O(1)。
不需要额外的空间。
如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。