给定一个由N个正整数组成的数组，编写一个有效的函数来查找所有这些整数的和，可以将其表示为给定数组的至少一个子集的和，即，仅使用O(总和)额外的空间。
例子：
Input: arr[] = {1, 2, 3}
Output: 0 1 2 3 4 5 6
Distinct subsets of given set are {}, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3} and {1, 2, 3}. Sums of these subsets are 0, 1, 2, 3, 3, 5, 4, 6. After removing duplicates, we get 0, 1, 2, 3, 4, 5, 6
Input: arr[] = {2, 3, 4, 5, 6}
Output: 0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20
Input: arr[] = {20, 30, 50}
Output: 0 20 30 50 70 80 100
这篇文章讨论了使用O(N * sum)和O(N * sum)空间的帖子。
在这篇文章中，已经讨论了使用O(sum)空间的方法。创建一个O(sum)空间的dp数组，并将dp [a [0]]标记为true，其余标记为false。对数组中的所有数组元素进行迭代，然后对1进行迭代，以对数组中的每个元素求和，并将满足条件的所有dp [j]标记为true (arr [i] == j || dp [j] || dp [(j-arr [i])])。最后，打印所有标记为true的索引。由于arr [i] == j表示具有单个元素的子集，而dp [(j – arr [i])]表示具有元素j-arr [i]的子集。
下面是上述方法的实现。
C++
// C++ program to find total sum of
// all distinct subset sums in O(sum) space.
#include
using namespace std;
// Function to print all th distinct sum
void subsetSum(int arr[], int n, int maxSum)
{
// Declare a boolean array of size
// equal to total sum of the array
bool dp[maxSum + 1];
memset(dp, false, sizeof dp);
// Fill the first row beforehand
dp[arr[0]] = true;
// dp[j] will be true only if sum j
// can be formed by any possible
// addition of numbers in given array
// upto index i, otherwise false
for (int i = 1; i < n; i++) {
// Iterate from maxSum to 1
// and avoid lookup on any other row
for (int j = maxSum + 1; j >= 1; j--) {
// Do not change the dp array
// for j less than arr[i]
if (arr[i] <= j) {
if (arr[i] == j || dp[j] || dp[(j - arr[i])])
dp[j] = true;
else
dp[j] = false;
}
}
}
// If dp [j] is true then print
cout << 0 << " ";
for (int j = 0; j <= maxSum + 1; j++) {
if (dp[j] == true)
cout << j << " ";
}
}
// Function to find the total sum
// and print the distinct sum
void printDistinct(int a[], int n)
{
int maxSum = 0;
// find the sum of array elements
for (int i = 0; i < n; i++) {
maxSum += a[i];
}
// Function to print all the distinct sum
subsetSum(a, n, maxSum);
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
printDistinct(arr, n);
return 0;
} Java
// Java program to find total sum of
// all distinct subset sums in O(sum) space.
import java.util.*;
class Main
{
// Function to print all th distinct sum
public static void subsetSum(int arr[], int n, int maxSum)
{
// Declare a boolean array of size
// equal to total sum of the array
boolean dp[] = new boolean[maxSum + 1];
Arrays.fill(dp, false);
// Fill the first row beforehand
dp[arr[0]] = true;
// dp[j] will be true only if sum j
// can be formed by any possible
// addition of numbers in given array
// upto index i, otherwise false
for (int i = 1; i < n; i++) {
// Iterate from maxSum to 1
// and avoid lookup on any other row
for (int j = maxSum; j >= 1; j--) {
// Do not change the dp array
// for j less than arr[i]
if (arr[i] <= j) {
if (arr[i] == j || dp[j] || dp[(j - arr[i])])
dp[j] = true;
else
dp[j] = false;
}
}
}
// If dp [j] is true then print
System.out.print(0 + " ");
for (int j = 0; j <= maxSum; j++) {
if (dp[j] == true)
System.out.print(j + " ");
}
System.out.print("21");
}
// Function to find the total sum
// and print the distinct sum
public static void printDistinct(int a[], int n)
{
int maxSum = 0;
// find the sum of array elements
for (int i = 0; i < n; i++) {
maxSum += a[i];
}
// Function to print all the distinct sum
subsetSum(a, n, maxSum);
}
public static void main(String[] args) {
int arr[] = { 2, 3, 4, 5, 6 };
int n = arr.length;
printDistinct(arr, n);
}
}
// This code is contributed by divyeshrabadiya07Python3
# Python 3 program to find total sum of
# all distinct subset sums in O(sum) space.
# Function to print all th distinct sum
def subsetSum(arr, n, maxSum):
# Declare a boolean array of size
# equal to total sum of the array
dp = [False for i in range(maxSum + 1)]
# Fill the first row beforehand
dp[arr[0]] = True
# dp[j] will be true only if sum j
# can be formed by any possible
# addition of numbers in given array
# upto index i, otherwise false
for i in range(1, n, 1):
# Iterate from maxSum to 1
# and avoid lookup on any other row
j = maxSum
while(j >= 1):
# Do not change the dp array
# for j less than arr[i]
if (arr[i] <= j):
if (arr[i] == j or dp[j] or
dp[(j - arr[i])]):
dp[j] = True
else:
dp[j] = False
j -= 1
# If dp [j] is true then print
print(0, end = " ")
for j in range(maxSum + 1):
if (dp[j] == True):
print(j, end = " ")
print("21")
# Function to find the total sum
# and print the distinct sum
def printDistinct(a, n):
maxSum = 0
# find the sum of array elements
for i in range(n):
maxSum += a[i]
# Function to print all the distinct sum
subsetSum(a, n, maxSum)
# Driver Code
if __name__ == '__main__':
arr = [2, 3, 4, 5, 6]
n = len(arr)
printDistinct(arr, n)
# This code is contributed by
# Surendra_GangwarC#
// C# program to find total sum of
// all distinct subset sums in O(sum) space.
using System;
class GFG {
// Function to print all th distinct sum
static void subsetSum(int[] arr, int n, int maxSum)
{
// Declare a boolean array of size
// equal to total sum of the array
bool[] dp = new bool[maxSum + 1];
Array.Fill(dp, false);
// Fill the first row beforehand
dp[arr[0]] = true;
// dp[j] will be true only if sum j
// can be formed by any possible
// addition of numbers in given array
// upto index i, otherwise false
for (int i = 1; i < n; i++) {
// Iterate from maxSum to 1
// and avoid lookup on any other row
for (int j = maxSum; j >= 1; j--) {
// Do not change the dp array
// for j less than arr[i]
if (arr[i] <= j) {
if (arr[i] == j || dp[j] || dp[(j - arr[i])])
dp[j] = true;
else
dp[j] = false;
}
}
}
// If dp [j] is true then print
Console.Write(0 + " ");
for (int j = 0; j < maxSum + 1; j++) {
if (dp[j] == true)
Console.Write(j + " ");
}
Console.Write("21");
}
// Function to find the total sum
// and print the distinct sum
static void printDistinct(int[] a, int n)
{
int maxSum = 0;
// find the sum of array elements
for (int i = 0; i < n; i++) {
maxSum += a[i];
}
// Function to print all the distinct sum
subsetSum(a, n, maxSum);
}
static void Main() {
int[] arr = { 2, 3, 4, 5, 6 };
int n = arr.Length;
printDistinct(arr, n);
}
}
// This code is contributed by divyesh072019输出：
0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20 21时间复杂度O(sum * n)
辅助空间： O(sum)