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
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)
}
其他回答
PHP版本,灵感来自Keith Beller的c#版本。
bala的PHP版本不适合我,因为我不需要对数字进行分组。我想要一个更简单的实现,只有一个目标值和一个数字池。这个函数也会删除任何重复的条目。
编辑25/10/2021:添加精度参数以支持浮点数(现在需要bcmath扩展)。
/**
* Calculates a subset sum: finds out which combinations of numbers
* from the numbers array can be added together to come to the target
* number.
*
* Returns an indexed array with arrays of number combinations.
*
* Example:
*
* <pre>
* $matches = subset_sum(array(5,10,7,3,20), 25);
* </pre>
*
* Returns:
*
* <pre>
* Array
* (
* [0] => Array
* (
* [0] => 3
* [1] => 5
* [2] => 7
* [3] => 10
* )
* [1] => Array
* (
* [0] => 5
* [1] => 20
* )
* )
* </pre>
*
* @param number[] $numbers
* @param number $target
* @param array $part
* @param int $precision
* @return array[number[]]
*/
function subset_sum($numbers, $target, $precision=0, $part=null)
{
// we assume that an empty $part variable means this
// is the top level call.
$toplevel = false;
if($part === null) {
$toplevel = true;
$part = array();
}
$s = 0;
foreach($part as $x)
{
$s = $s + $x;
}
// we have found a match!
if(bccomp((string) $s, (string) $target, $precision) === 0)
{
sort($part); // ensure the numbers are always sorted
return array(implode('|', $part));
}
// gone too far, break off
if($s >= $target)
{
return null;
}
$matches = array();
$totalNumbers = count($numbers);
for($i=0; $i < $totalNumbers; $i++)
{
$remaining = array();
$n = $numbers[$i];
for($j = $i+1; $j < $totalNumbers; $j++)
{
$remaining[] = $numbers[$j];
}
$part_rec = $part;
$part_rec[] = $n;
$result = subset_sum($remaining, $target, $precision, $part_rec);
if($result)
{
$matches = array_merge($matches, $result);
}
}
if(!$toplevel)
{
return $matches;
}
// this is the top level function call: we have to
// prepare the final result value by stripping any
// duplicate results.
$matches = array_unique($matches);
$result = array();
foreach($matches as $entry)
{
$result[] = explode('|', $entry);
}
return $result;
}
例子:
$result = subset_sum(array(5, 10, 7, 3, 20), 25);
这将返回一个包含两个数字组合数组的索引数组:
3, 5, 7, 10
5, 20
浮点数示例:
// Specify the precision in the third argument
$result = subset_sum(array(0.40, 0.03, 0.05), 0.45, 2);
这将返回一个匹配项:
0.40, 0.05
Java解决方案的Swift 3转换(by @JeremyThompson)
protocol _IntType { }
extension Int: _IntType {}
extension Array where Element: _IntType {
func subsets(to: Int) -> [[Element]]? {
func sum_up_recursive(_ numbers: [Element], _ target: Int, _ partial: [Element], _ solution: inout [[Element]]) {
var sum: Int = 0
for x in partial {
sum += x as! Int
}
if sum == target {
solution.append(partial)
}
guard sum < target else {
return
}
for i in stride(from: 0, to: numbers.count, by: 1) {
var remaining = [Element]()
for j in stride(from: i + 1, to: numbers.count, by: 1) {
remaining.append(numbers[j])
}
var partial_rec = [Element](partial)
partial_rec.append(numbers[i])
sum_up_recursive(remaining, target, partial_rec, &solution)
}
}
var solutions = [[Element]]()
sum_up_recursive(self, to, [Element](), &solutions)
return solutions.count > 0 ? solutions : nil
}
}
用法:
let numbers = [3, 9, 8, 4, 5, 7, 10]
if let solution = numbers.subsets(to: 15) {
print(solution) // output: [[3, 8, 4], [3, 5, 7], [8, 7], [5, 10]]
} else {
print("not possible")
}
import java.util.*;
public class Main{
int recursionDepth = 0;
private int[][] memo;
public static void main(String []args){
int[] nums = new int[] {5,2,4,3,1};
int N = nums.length;
Main main = new Main();
main.memo = new int[N+1][N+1];
main._findCombo(0, N-1,nums, 8, 0, new LinkedList() );
System.out.println(main.recursionDepth);
}
private void _findCombo(
int from,
int to,
int[] nums,
int targetSum,
int currentSum,
LinkedList<Integer> list){
if(memo[from][to] != 0) {
currentSum = currentSum + memo[from][to];
}
if(currentSum > targetSum) {
return;
}
if(currentSum == targetSum) {
System.out.println("Found - " +list);
return;
}
recursionDepth++;
for(int i= from ; i <= to; i++){
list.add(nums[i]);
memo[from][i] = currentSum + nums[i];
_findCombo(i+1, to,nums, targetSum, memo[from][i], list);
list.removeLast();
}
}
}
Thank you.. ephemient
我已经将上述逻辑从python转换为php..
<?php
$data = array(array(2,3,5,10,15),array(4,6,23,15,12),array(23,34,12,1,5));
$maxsum = 25;
print_r(bestsum($data,$maxsum)); //function call
function bestsum($data,$maxsum)
{
$res = array_fill(0, $maxsum + 1, '0');
$res[0] = array(); //base case
foreach($data as $group)
{
$new_res = $res; //copy res
foreach($group as $ele)
{
for($i=0;$i<($maxsum-$ele+1);$i++)
{
if($res[$i] != 0)
{
$ele_index = $i+$ele;
$new_res[$ele_index] = $res[$i];
$new_res[$ele_index][] = $ele;
}
}
}
$res = $new_res;
}
for($i=$maxsum;$i>0;$i--)
{
if($res[$i]!=0)
{
return $res[$i];
break;
}
}
return array();
}
?>
我想我应该用这个问题的答案,但我不能,所以这是我的答案。它使用的是《计算机程序的结构和解释》中答案的修改版本。我认为这是一个更好的递归解,应该更能取悦纯粹主义者。
我的答案是用Scala(如果我的Scala很烂,我很抱歉,我刚刚开始学习)。findsumcombination的疯狂之处在于对递归的原始列表进行排序和惟一,以防止欺骗。
def findSumCombinations(target: Int, numbers: List[Int]): Int = {
cc(target, numbers.distinct.sortWith(_ < _), List())
}
def cc(target: Int, numbers: List[Int], solution: List[Int]): Int = {
if (target == 0) {println(solution); 1 }
else if (target < 0 || numbers.length == 0) 0
else
cc(target, numbers.tail, solution)
+ cc(target - numbers.head, numbers, numbers.head :: solution)
}
使用它:
> findSumCombinations(12345, List(1,5,22,15,0,..))
* Prints a whole heap of lists that will sum to the target *