你如何从给定的N个数字中测试所有可能的加法组合,使它们加起来得到给定的最终数字?
一个简单的例子:
要添加的数字集:N ={1,5,22,15,0,…} 期望结果:12345
你如何从给定的N个数字中测试所有可能的加法组合,使它们加起来得到给定的最终数字?
一个简单的例子:
要添加的数字集:N ={1,5,22,15,0,…} 期望结果:12345
当前回答
下面是一个更好的版本,具有更好的输出格式和c++ 11特性:
void subset_sum_rec(std::vector<int> & nums, const int & target, std::vector<int> & partialNums)
{
int currentSum = std::accumulate(partialNums.begin(), partialNums.end(), 0);
if (currentSum > target)
return;
if (currentSum == target)
{
std::cout << "sum([";
for (auto it = partialNums.begin(); it != std::prev(partialNums.end()); ++it)
cout << *it << ",";
cout << *std::prev(partialNums.end());
std::cout << "])=" << target << std::endl;
}
for (auto it = nums.begin(); it != nums.end(); ++it)
{
std::vector<int> remaining;
for (auto it2 = std::next(it); it2 != nums.end(); ++it2)
remaining.push_back(*it2);
std::vector<int> partial = partialNums;
partial.push_back(*it);
subset_sum_rec(remaining, target, partial);
}
}
其他回答
另一个python解决方案是使用itertools.combination模块,如下所示:
#!/usr/local/bin/python
from itertools import combinations
def find_sum_in_list(numbers, target):
results = []
for x in range(len(numbers)):
results.extend(
[
combo for combo in combinations(numbers ,x)
if sum(combo) == target
]
)
print results
if __name__ == "__main__":
find_sum_in_list([3,9,8,4,5,7,10], 15)
输出:[(8,7),(5,10),(3,8,4),(3,5,7)]
我将c#示例移植到Objective-c,并没有在响应中看到它:
//Usage
NSMutableArray* numberList = [[NSMutableArray alloc] init];
NSMutableArray* partial = [[NSMutableArray alloc] init];
int target = 16;
for( int i = 1; i<target; i++ )
{ [numberList addObject:@(i)]; }
[self findSums:numberList target:target part:partial];
//*******************************************************************
// Finds combinations of numbers that add up to target recursively
//*******************************************************************
-(void)findSums:(NSMutableArray*)numbers target:(int)target part:(NSMutableArray*)partial
{
int s = 0;
for (NSNumber* x in partial)
{ s += [x intValue]; }
if (s == target)
{ NSLog(@"Sum[%@]", partial); }
if (s >= target)
{ return; }
for (int i = 0;i < [numbers count];i++ )
{
int n = [numbers[i] intValue];
NSMutableArray* remaining = [[NSMutableArray alloc] init];
for (int j = i + 1; j < [numbers count];j++)
{ [remaining addObject:@([numbers[j] intValue])]; }
NSMutableArray* partRec = [[NSMutableArray alloc] initWithArray:partial];
[partRec addObject:@(n)];
[self findSums:remaining target:target part:partRec];
}
}
下面是一个Java版本,它非常适合小N和非常大的目标和,当复杂度O(t*N)(动态解)大于指数算法时。我的版本在中间攻击中使用了一个meet,并进行了一些调整,以降低复杂度,从经典的naive O(n*2^n)降低到O(2^(n/2))。
如果你想在32到64个元素之间的集合中使用这种方法,你应该将表示step函数中当前子集的int改为long,尽管随着集合大小的增加,性能显然会急剧下降。如果你想对一个有奇数个元素的集合使用这个,你应该给这个集合加上一个0,使它成为偶数。
import java.util.ArrayList;
import java.util.List;
public class SubsetSumMiddleAttack {
static final int target = 100000000;
static final int[] set = new int[]{ ... };
static List<Subset> evens = new ArrayList<>();
static List<Subset> odds = new ArrayList<>();
static int[][] split(int[] superSet) {
int[][] ret = new int[2][superSet.length / 2];
for (int i = 0; i < superSet.length; i++) ret[i % 2][i / 2] = superSet[i];
return ret;
}
static void step(int[] superSet, List<Subset> accumulator, int subset, int sum, int counter) {
accumulator.add(new Subset(subset, sum));
if (counter != superSet.length) {
step(superSet, accumulator, subset + (1 << counter), sum + superSet[counter], counter + 1);
step(superSet, accumulator, subset, sum, counter + 1);
}
}
static void printSubset(Subset e, Subset o) {
String ret = "";
for (int i = 0; i < 32; i++) {
if (i % 2 == 0) {
if ((1 & (e.subset >> (i / 2))) == 1) ret += " + " + set[i];
}
else {
if ((1 & (o.subset >> (i / 2))) == 1) ret += " + " + set[i];
}
}
if (ret.startsWith(" ")) ret = ret.substring(3) + " = " + (e.sum + o.sum);
System.out.println(ret);
}
public static void main(String[] args) {
int[][] superSets = split(set);
step(superSets[0], evens, 0,0,0);
step(superSets[1], odds, 0,0,0);
for (Subset e : evens) {
for (Subset o : odds) {
if (e.sum + o.sum == target) printSubset(e, o);
}
}
}
}
class Subset {
int subset;
int sum;
Subset(int subset, int sum) {
this.subset = subset;
this.sum = sum;
}
}
在Haskell:
filter ((==) 12345 . sum) $ subsequences [1,5,22,15,0,..]
J:
(]#~12345=+/@>)(]<@#~[:#:@i.2^#)1 5 22 15 0 ...
正如您可能注意到的,两者都采用相同的方法,并将问题分为两部分:生成幂集的每个成员,并检查每个成员与目标的和。
还有其他的解决方案,但这是最直接的。
在这两种方法中,你是否需要帮助,或者找到另一种方法?
Excel VBA版本如下。我需要在VBA中实现这一点(不是我的偏好,不要评判我!),并使用本页上的答案作为方法。我上传以防其他人也需要VBA版本。
Option Explicit
Public Sub SumTarget()
Dim numbers(0 To 6) As Long
Dim target As Long
target = 15
numbers(0) = 3: numbers(1) = 9: numbers(2) = 8: numbers(3) = 4: numbers(4) = 5
numbers(5) = 7: numbers(6) = 10
Call SumUpTarget(numbers, target)
End Sub
Public Sub SumUpTarget(numbers() As Long, target As Long)
Dim part() As Long
Call SumUpRecursive(numbers, target, part)
End Sub
Private Sub SumUpRecursive(numbers() As Long, target As Long, part() As Long)
Dim s As Long, i As Long, j As Long, num As Long
Dim remaining() As Long, partRec() As Long
s = SumArray(part)
If s = target Then Debug.Print "SUM ( " & ArrayToString(part) & " ) = " & target
If s >= target Then Exit Sub
If (Not Not numbers) <> 0 Then
For i = 0 To UBound(numbers)
Erase remaining()
num = numbers(i)
For j = i + 1 To UBound(numbers)
AddToArray remaining, numbers(j)
Next j
Erase partRec()
CopyArray partRec, part
AddToArray partRec, num
SumUpRecursive remaining, target, partRec
Next i
End If
End Sub
Private Function ArrayToString(x() As Long) As String
Dim n As Long, result As String
result = "{" & x(n)
For n = LBound(x) + 1 To UBound(x)
result = result & "," & x(n)
Next n
result = result & "}"
ArrayToString = result
End Function
Private Function SumArray(x() As Long) As Long
Dim n As Long
SumArray = 0
If (Not Not x) <> 0 Then
For n = LBound(x) To UBound(x)
SumArray = SumArray + x(n)
Next n
End If
End Function
Private Sub AddToArray(arr() As Long, x As Long)
If (Not Not arr) <> 0 Then
ReDim Preserve arr(0 To UBound(arr) + 1)
Else
ReDim Preserve arr(0 To 0)
End If
arr(UBound(arr)) = x
End Sub
Private Sub CopyArray(destination() As Long, source() As Long)
Dim n As Long
If (Not Not source) <> 0 Then
For n = 0 To UBound(source)
AddToArray destination, source(n)
Next n
End If
End Sub
输出(写入立即窗口)应该是:
SUM ( {3,8,4} ) = 15
SUM ( {3,5,7} ) = 15
SUM ( {8,7} ) = 15
SUM ( {5,10} ) = 15