一个 function 返回所有可能的子 arrays 总和为特定给定数字（不必是连续子数组）

Question

当我做一些数学运算时，我遇到了这个问题，我需要找到给定数字的所有可能总和。 这是一个示例输入和 output 以便更好地理解问题。 假设我们得到一个数组arr=[1,2,3,4,5]并且给定数字number=10那么我需要这样的东西，

def solution(arr,number):
    #you can code here

output： [1,2,3,4],[2,3,5],[1,5,4],[1,3,6],[1,6]

Answer 1

这里和这里都有很多有趣的答案和讨论。

这是一个快速（经过测试）的答案：

import itertools

def solution(arr, number):
    for i in range(len(arr)+1):
        for s in list(itertools.combinations(arr, i)):
            if sum(s) == number:
                yield list(s)


arr = [ 1, 2, 3, 4, 5 ]
number = 10

print(list(solution(arr, number)))

印刷：

[[1, 4, 5], [2, 3, 5], [1, 2, 3, 4]]

Answer 2

使用动态规划算法求解

def solve(arr, s, i = None, path = None, solutions = None):
  """Use Dynamic Programming to find subsequence of arr
     that sum to s
     
     Arguments:
         arr      - array
         i        - current index that will use or not for arr
         path     - current subsequence we are summing
        solutions - all solutions found
  """
  # Set defaults
  if i is None:
    i = len(arr) - 1
  if solutions is None:
    solutions = []
  if path is None:
    path = []

  # Base Cases
  if s == 0:
    solutions.append(path[:])
    return solutions

  if i < 0 or s < 0:
    return

  # Try with arr[i] in path
  if arr[i] <= s:
    solve(arr, s-arr[i], i-1, [arr[i]] + path, solutions)

  # Try with arr[i] not in path
  solve(arr, s, i-1, path, solutions)

  return solutions

测试

arr = [1, 2, 3, 4, 5]
print(solve(arr, 10))

Output

[[1, 4, 5], [2, 3, 5], [1, 2, 3, 4]]