给定一个数组 a，我们必须找到数组中存在的元素子集的最大乘积。最大乘积也可以是单个元素。
例子：
Input: a[] = { -1, -1, -2, 4, 3 }
Output: 24
Explanation : Maximum product will be ( -2 * -1 * 4 * 3 ) = 24
Input: a[] = { -1, 0 }
Output: 0
Explanation: 0(single element) is maximum product possible
Input: a[] = { 0, 0, 0 }
Output: 0一个简单的解决方案是生成所有子集，找到每个子集的乘积并返回最大乘积。
更好的解决方案是使用以下事实。
- 如果有偶数个负数且没有零，则结果只是所有的乘积
- 如果有奇数个负数且没有零，则结果是除绝对值最小的负整数之外的所有数的乘积。
- 如果有零，则结果是除这些零之外的所有零和一个例外情况的乘积。例外情况是当有一个负数且所有其他元素为 0 时。在这种情况下，结果为 0。
下面是上述方法的实现：
C++
// CPP program to find maximum product of
// a subset.
#include
using namespace std;
int maxProductSubset(int a[], int n)
{
if (n == 1)
return a[0];
// Find count of negative numbers, count
// of zeros, negative number
// with least absolute value
// and product of non-zero numbers
int max_neg = INT_MIN;
int count_neg = 0, count_zero = 0;
int prod = 1;
for (int i = 0; i < n; i++) {
// If number is 0, we don't
// multiply it with product.
if (a[i] == 0) {
count_zero++;
continue;
}
// Count negatives and keep
// track of negative number
// with least absolute value
if (a[i] < 0) {
count_neg++;
max_neg = max(max_neg, a[i]);
}
prod = prod * a[i];
}
// If there are all zeros
if (count_zero == n)
return 0;
// If there are odd number of
// negative numbers
if (count_neg & 1) {
// Exceptional case: There is only
// negative and all other are zeros
if (count_neg == 1 &&
count_zero > 0 &&
count_zero + count_neg == n)
return 0;
// Otherwise result is product of
// all non-zeros divided by
//negative number with
// least absolute value
prod = prod / max_neg;
}
return prod;
}
// Driver Code
int main()
{
int a[] = { -1, -1, -2, 4, 3 };
int n = sizeof(a) / sizeof(a[0]);
cout << maxProductSubset(a, n);
return 0;
} Java
// Java program to find maximum product of
// a subset.
class GFG {
static int maxProductSubset(int a[], int n) {
if (n == 1) {
return a[0];
}
// Find count of negative numbers, count
// of zeros, negative number
// with least absolute value
// and product of non-zero numbers
int max_neg = Integer.MIN_VALUE;
int count_neg = 0, count_zero = 0;
int prod = 1;
for (int i = 0; i < n; i++) {
// If number is 0, we don't
// multiply it with product.
if (a[i] == 0) {
count_zero++;
continue;
}
// Count negatives and keep
// track of negative number
// with least absolute value.
if (a[i] < 0) {
count_neg++;
max_neg = Math.max(max_neg, a[i]);
}
prod = prod * a[i];
}
// If there are all zeros
if (count_zero == n) {
return 0;
}
// If there are odd number of
// negative numbers
if (count_neg % 2 == 1) {
// Exceptional case: There is only
// negative and all other are zeros
if (count_neg == 1
&& count_zero > 0
&& count_zero + count_neg == n) {
return 0;
}
// Otherwise result is product of
// all non-zeros divided by
//negative number with
// least absolute value.
prod = prod / max_neg;
}
return prod;
}
// Driver Code
public static void main(String[] args) {
int a[] = {-1, -1, -2, 4, 3};
int n = a.length;
System.out.println(maxProductSubset(a, n));
}
}
/* This JAVA code is contributed by Rajput-Ji*/Python3
# Python3 program to find maximum product
# of a subset.
def maxProductSubset(a, n):
if n == 1:
return a[0]
# Find count of negative numbers, count
# of zeros, negative number
# with least absolute value
# and product of non-zero numbers
max_neg = -999999999999
count_neg = 0
count_zero = 0
prod = 1
for i in range(n):
# If number is 0, we don't
# multiply it with product.
if a[i] == 0:
count_zero += 1
continue
# Count negatives and keep
# track of negative number
# with least absolute value.
if a[i] < 0:
count_neg += 1
max_neg = max(max_neg, a[i])
prod = prod * a[i]
# If there are all zeros
if count_zero == n:
return 0
# If there are odd number of
# negative numbers
if count_neg & 1:
# Exceptional case: There is only
# negative and all other are zeros
if (count_neg == 1 and count_zero > 0 and
count_zero + count_neg == n):
return 0
# Otherwise result is product of
# all non-zeros divided
# by negative number
# with least absolute value
prod = int(prod / max_neg)
return prod
# Driver Code
if __name__ == '__main__':
a = [ -1, -1, -2, 4, 3 ]
n = len(a)
print(maxProductSubset(a, n))
# This code is contributed by PranchalKC#
// C# Java program to find maximum
// product of a subset.
using System;
class GFG
{
static int maxProductSubset(int []a,
int n)
{
if (n == 1)
{
return a[0];
}
// Find count of negative numbers,
// count of zeros, negative number with
// least absolute value and product of
// non-zero numbers
int max_neg = int.MinValue;
int count_neg = 0, count_zero = 0;
int prod = 1;
for (int i = 0; i < n; i++)
{
// If number is 0, we don't
// multiply it with product.
if (a[i] == 0)
{
count_zero++;
continue;
}
// Count negatives and keep
// track of negative number with
// least absolute value.
if (a[i] < 0)
{
count_neg++;
max_neg = Math.Max(max_neg, a[i]);
}
prod = prod * a[i];
}
// If there are all zeros
if (count_zero == n)
{
return 0;
}
// If there are odd number of
// negative numbers
if (count_neg % 2 == 1)
{
// Exceptional case: There is only
// negative and all other are zeros
if (count_neg == 1 && count_zero > 0 &&
count_zero + count_neg == n)
{
return 0;
}
// Otherwise result is product of
// all non-zeros divided by negative
// number with least absolute value.
prod = prod / max_neg;
}
return prod;
}
// Driver code
public static void Main()
{
int []a = {-1, -1, -2, 4, 3};
int n = a.Length;
Console.Write(maxProductSubset(a, n));
}
}
// This code is contributed by Rajput-JiPHP
0 &&
$count_zero + $count_neg == $n)
return 0;
// Otherwise result is product of
// all non-zeros divided by negative
// number with least absolute value.
$prod = $prod / $max_neg;
}
return $prod;
}
// Driver Code
$a = array(-1, -1, -2, 4, 3 );
$n = sizeof($a);
echo maxProductSubset($a, $n);
// This code is contributed
// by Akanksha Rai
?>Javascript
输出
24时间复杂度： O(n)
辅助空间： O(1)
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