获得所有可能的总和,加起来给定的数字

Question

我正在为android制作一个数学应用程序.在其中一个字段中,用户可以输入int(无数字且高于0).这个想法是获得所有可能的和,使得这个int,没有双打(在这种情况下4 + 1 == 1 + 4).唯一知道的是这一个int.

例如:

假设用户输入4,我希望应用程序返回:

• 4
• 3 + 1
• 2 + 2
• 2 + 1 + 1
• 1 + 1 + 1 + 1

显然4 == 4所以也应该加上.关于我应该如何做这个的任何建议？

Answer 1

这是一个简单的算法,声称这样做

来自:http://introcs.cs.princeton.edu/java/23recursion/Partition.java.html

public class Partition { 

    public static void partition(int n) {
        partition(n, n, "");
    }
    public static void partition(int n, int max, String prefix) {
        if (n == 0) {
            StdOut.println(prefix);
            return;
        }

        for (int i = Math.min(max, n); i >= 1; i--) {
            partition(n-i, i, prefix + " " + i);
        }
    }


    public static void main(String[] args) { 
        int N = Integer.parseInt(args[0]);
        partition(N);
    }

}

Answer 2

有一个简短而优雅的递归解决方案来生成它们,但以下可能更容易使用并在现有代码中实现:

import java.util.*;

public class SumIterator implements Iterator<List<Integer>>, Iterable<List<Integer>> {

  // keeps track of all sums that have been generated already
  private Set<List<Integer>> generated;

  // holds all sums that haven't been returned by `next()`
  private Stack<List<Integer>> sums;

  public SumIterator(int n) {

    // first a sanity check...
    if(n < 1) {
      throw new RuntimeException("'n' must be >= 1");
    }

    generated = new HashSet<List<Integer>>();
    sums = new Stack<List<Integer>>();

    // create and add the "last" sum of size `n`: [1, 1, 1, ... , 1]
    List<Integer> last = new ArrayList<Integer>();
    for(int i = 0; i < n; i++) {
      last.add(1);
    }
    add(last);

    // add the first sum of size 1: [n]
    add(Arrays.asList(n));
  }

  private void add(List<Integer> sum) {
    if(generated.add(sum)) {
      // only push the sum on the stack if it hasn't been generated before
      sums.push(sum);
    }
  }

  @Override
  public boolean hasNext() {
    return !sums.isEmpty();
  }

  @Override
  public Iterator<List<Integer>> iterator() {
    return this;
  }

  @Override
  public List<Integer> next() {
    List<Integer> sum = sums.pop();                         // get the next sum from the stack
    for(int i = sum.size() - 1; i >= 0; i--) {              // loop from right to left
      int n = sum.get(i);                                   //   get the i-th number
      if(n > 1) {                                           //   if the i-th number is more than 1
        for(int j = n-1; j > n/2; j--) {                    //     if the i-th number is 10, loop from 9 to 5
          List<Integer> copy = new ArrayList<Integer>(sum); //       create a copy of the current sum
          copy.remove(i);                                   //       remove the i-th number
          copy.add(i, j);                                   //       insert `j` where the i-th number was
          copy.add(i + 1, n-j);                             //       insert `n-j` next to `j`
          add(copy);                                        //       add this new sum to the stack
        }                                                   //     
        break;                                              //   stop looping any further
      }                                                     
    }
    return sum;
  }

  @Override
  public void remove() {
    throw new UnsupportedOperationException();
  }
}

你可以像这样使用它:

int n = 10;
for(List<Integer> sum : new SumIterator(n)) {
  System.out.println(n + " = " + sum);
}

哪个会打印:

10 = [10]
10 = [6, 4]
10 = [6, 3, 1]
10 = [6, 2, 1, 1]
10 = [7, 3]
10 = [7, 2, 1]
10 = [8, 2]
10 = [9, 1]
10 = [5, 4, 1]
10 = [5, 3, 1, 1]
10 = [5, 2, 1, 1, 1]
10 = [8, 1, 1]
10 = [7, 1, 1, 1]
10 = [4, 3, 1, 1, 1]
10 = [4, 2, 1, 1, 1, 1]
10 = [6, 1, 1, 1, 1]
10 = [5, 1, 1, 1, 1, 1]
10 = [3, 2, 1, 1, 1, 1, 1]
10 = [4, 1, 1, 1, 1, 1, 1]
10 = [3, 1, 1, 1, 1, 1, 1, 1]
10 = [2, 1, 1, 1, 1, 1, 1, 1, 1]
10 = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]