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;
}
}
其他回答
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;
}
}
我不喜欢上面看到的Javascript解决方案。下面是我使用部分应用、闭包和递归构建的一个:
好的，我主要关心的是，如果组合数组能满足目标要求，希望这样你就能找到剩下的组合了
这里只需要设置目标并传递组合数组。
function main() {
const target = 10
const getPermutationThatSumT = setTarget(target)
const permutation = getPermutationThatSumT([1, 4, 2, 5, 6, 7])
console.log( permutation );
}
我提出的当前实现
function setTarget(target) {
let partial = [];
return function permute(input) {
let i, removed;
for (i = 0; i < input.length; i++) {
removed = input.splice(i, 1)[0];
partial.push(removed);
const sum = partial.reduce((a, b) => a + b)
if (sum === target) return partial.slice()
if (sum < target) permute(input)
input.splice(i, 0, removed);
partial.pop();
}
return null
};
}
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];
}
}
Java非递归版本，简单地添加元素并在可能的值之间重新分配它们。0被忽略，适用于固定的列表(给定的是您可以使用的)或可重复的数字列表。
import java.util.*;
public class TestCombinations {
public static void main(String[] args) {
ArrayList<Integer> numbers = new ArrayList<>(Arrays.asList(0, 1, 2, 2, 5, 10, 20));
LinkedHashSet<Integer> targets = new LinkedHashSet<Integer>() {{
add(4);
add(10);
add(25);
}};
System.out.println("## each element can appear as many times as needed");
for (Integer target: targets) {
Combinations combinations = new Combinations(numbers, target, true);
combinations.calculateCombinations();
for (String solution: combinations.getCombinations()) {
System.out.println(solution);
}
}
System.out.println("## each element can appear only once");
for (Integer target: targets) {
Combinations combinations = new Combinations(numbers, target, false);
combinations.calculateCombinations();
for (String solution: combinations.getCombinations()) {
System.out.println(solution);
}
}
}
public static class Combinations {
private boolean allowRepetitions;
private int[] repetitions;
private ArrayList<Integer> numbers;
private Integer target;
private Integer sum;
private boolean hasNext;
private Set<String> combinations;
/**
* Constructor.
*
* @param numbers Numbers that can be used to calculate the sum.
* @param target Target value for sum.
*/
public Combinations(ArrayList<Integer> numbers, Integer target) {
this(numbers, target, true);
}
/**
* Constructor.
*
* @param numbers Numbers that can be used to calculate the sum.
* @param target Target value for sum.
*/
public Combinations(ArrayList<Integer> numbers, Integer target, boolean allowRepetitions) {
this.allowRepetitions = allowRepetitions;
if (this.allowRepetitions) {
Set<Integer> numbersSet = new HashSet<>(numbers);
this.numbers = new ArrayList<>(numbersSet);
} else {
this.numbers = numbers;
}
this.numbers.removeAll(Arrays.asList(0));
Collections.sort(this.numbers);
this.target = target;
this.repetitions = new int[this.numbers.size()];
this.combinations = new LinkedHashSet<>();
this.sum = 0;
if (this.repetitions.length > 0)
this.hasNext = true;
else
this.hasNext = false;
}
/**
* Calculate and return the sum of the current combination.
*
* @return The sum.
*/
private Integer calculateSum() {
this.sum = 0;
for (int i = 0; i < repetitions.length; ++i) {
this.sum += repetitions[i] * numbers.get(i);
}
return this.sum;
}
/**
* Redistribute picks when only one of each number is allowed in the sum.
*/
private void redistribute() {
for (int i = 1; i < this.repetitions.length; ++i) {
if (this.repetitions[i - 1] > 1) {
this.repetitions[i - 1] = 0;
this.repetitions[i] += 1;
}
}
if (this.repetitions[this.repetitions.length - 1] > 1)
this.repetitions[this.repetitions.length - 1] = 0;
}
/**
* Get the sum of the next combination. When 0 is returned, there's no other combinations to check.
*
* @return The sum.
*/
private Integer next() {
if (this.hasNext && this.repetitions.length > 0) {
this.repetitions[0] += 1;
if (!this.allowRepetitions)
this.redistribute();
this.calculateSum();
for (int i = 0; i < this.repetitions.length && this.sum != 0; ++i) {
if (this.sum > this.target) {
this.repetitions[i] = 0;
if (i + 1 < this.repetitions.length) {
this.repetitions[i + 1] += 1;
if (!this.allowRepetitions)
this.redistribute();
}
this.calculateSum();
}
}
if (this.sum.compareTo(0) == 0)
this.hasNext = false;
}
return this.sum;
}
/**
* Calculate all combinations whose sum equals target.
*/
public void calculateCombinations() {
while (this.hasNext) {
if (this.next().compareTo(target) == 0)
this.combinations.add(this.toString());
}
}
/**
* Return all combinations whose sum equals target.
*
* @return Combinations as a set of strings.
*/
public Set<String> getCombinations() {
return this.combinations;
}
@Override
public String toString() {
StringBuilder stringBuilder = new StringBuilder("" + sum + ": ");
for (int i = 0; i < repetitions.length; ++i) {
for (int j = 0; j < repetitions[i]; ++j) {
stringBuilder.append(numbers.get(i) + " ");
}
}
return stringBuilder.toString();
}
}
}
样例输入:
numbers: 0, 1, 2, 2, 5, 10, 20
targets: 4, 10, 25
样例输出:
## each element can appear as many times as needed
4: 1 1 1 1
4: 1 1 2
4: 2 2
10: 1 1 1 1 1 1 1 1 1 1
10: 1 1 1 1 1 1 1 1 2
10: 1 1 1 1 1 1 2 2
10: 1 1 1 1 2 2 2
10: 1 1 2 2 2 2
10: 2 2 2 2 2
10: 1 1 1 1 1 5
10: 1 1 1 2 5
10: 1 2 2 5
10: 5 5
10: 10
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2
25: 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
25: 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2
25: 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
25: 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2
25: 1 1 1 2 2 2 2 2 2 2 2 2 2 2
25: 1 2 2 2 2 2 2 2 2 2 2 2 2
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 5
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 5
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 5
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 5
25: 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 5
25: 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 5
25: 1 1 1 1 1 1 1 1 2 2 2 2 2 2 5
25: 1 1 1 1 1 1 2 2 2 2 2 2 2 5
25: 1 1 1 1 2 2 2 2 2 2 2 2 5
25: 1 1 2 2 2 2 2 2 2 2 2 5
25: 2 2 2 2 2 2 2 2 2 2 5
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 5 5
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 2 5 5
25: 1 1 1 1 1 1 1 1 1 1 1 2 2 5 5
25: 1 1 1 1 1 1 1 1 1 2 2 2 5 5
25: 1 1 1 1 1 1 1 2 2 2 2 5 5
25: 1 1 1 1 1 2 2 2 2 2 5 5
25: 1 1 1 2 2 2 2 2 2 5 5
25: 1 2 2 2 2 2 2 2 5 5
25: 1 1 1 1 1 1 1 1 1 1 5 5 5
25: 1 1 1 1 1 1 1 1 2 5 5 5
25: 1 1 1 1 1 1 2 2 5 5 5
25: 1 1 1 1 2 2 2 5 5 5
25: 1 1 2 2 2 2 5 5 5
25: 2 2 2 2 2 5 5 5
25: 1 1 1 1 1 5 5 5 5
25: 1 1 1 2 5 5 5 5
25: 1 2 2 5 5 5 5
25: 5 5 5 5 5
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10
25: 1 1 1 1 1 1 1 1 1 1 1 1 1 2 10
25: 1 1 1 1 1 1 1 1 1 1 1 2 2 10
25: 1 1 1 1 1 1 1 1 1 2 2 2 10
25: 1 1 1 1 1 1 1 2 2 2 2 10
25: 1 1 1 1 1 2 2 2 2 2 10
25: 1 1 1 2 2 2 2 2 2 10
25: 1 2 2 2 2 2 2 2 10
25: 1 1 1 1 1 1 1 1 1 1 5 10
25: 1 1 1 1 1 1 1 1 2 5 10
25: 1 1 1 1 1 1 2 2 5 10
25: 1 1 1 1 2 2 2 5 10
25: 1 1 2 2 2 2 5 10
25: 2 2 2 2 2 5 10
25: 1 1 1 1 1 5 5 10
25: 1 1 1 2 5 5 10
25: 1 2 2 5 5 10
25: 5 5 5 10
25: 1 1 1 1 1 10 10
25: 1 1 1 2 10 10
25: 1 2 2 10 10
25: 5 10 10
25: 1 1 1 1 1 20
25: 1 1 1 2 20
25: 1 2 2 20
25: 5 20
## each element can appear only once
4: 2 2
10: 1 2 2 5
10: 10
25: 1 2 2 20
25: 5 20
我在做类似的scala作业。我想在这里发布我的解决方案:
def countChange(money: Int, coins: List[Int]): Int = {
def getCount(money: Int, remainingCoins: List[Int]): Int = {
if(money == 0 ) 1
else if(money < 0 || remainingCoins.isEmpty) 0
else
getCount(money, remainingCoins.tail) +
getCount(money - remainingCoins.head, remainingCoins)
}
if(money == 0 || coins.isEmpty) 0
else getCount(money, coins)
}
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