寻找多次存储数组元素的特定和的算法

def find_combinations(nums, target):
    result = []
    backtrack(nums, target, 0, [], 0, result)
    return result

def backtrack(nums, target, index, combination, current_sum, result):
    if current_sum == target:
        result.append(combination[:])
        return
    if current_sum > target or index >= len(nums):
        return

    for i in range(index, len(nums)):
        combination.append(nums[i])
        current_sum += nums[i]
        backtrack(nums, target, i, combination, current_sum, result)
        combination.pop()
        current_sum -= nums[i]

# 示例用法
nums = [1, 2, 3, 4, 5]
target = 5
result = find_combinations(nums, target)
print(result)