条件满足时停止 for 循环或 while 循环

Question

I am newby in programming. The goal is to count numbers in list in order, but loop has to stop when condition is met or close to it, but must not exceed it.

For example: list = [4,4,4,3,3], condition = 11 Expected output will be 4+4=8, because another 4 will exceed condition(4+4+4=12).

list = [4,4,4,3,3]
condition = 11
def for_fun(list, condition):
    sum_of_sizes = 0
    for i in list:
        sum_of_sizes += i
        if sum_of_sizes <= condition:
            break

    return sum_of_sizes

I find easier to work with for loops although while loop might be better for this task.

Answer 1

I suggest you check if your next sum_of_sizes is greater or equal than your condition and break if this is the case. If not, just proceed to add items from your list. Also, list is a built-in function, so please call your list something else like input_list:

input_list = [4,4,4,3,3]
condition = 11

def for_fun(input_list, condition):
    """A function that adds up numbers and breaks if the next iteration would exceed condition"""
    
    # initial sum of numbers
    sum_of_sizes = 0
    
    # iterate over your input list
    for i in input_list:

        # if the next sum of numbers would exceed condition, break
        if sum_of_sizes+i > condition:
            break

        # add up numbers
        sum_of_sizes += i

    # return the sum at the end
    return sum_of_sizes

# call and print your function, hand over your variables input_list and condition
print(for_fun(input_list=input_list, condition=condition))

Answer 2

To add a solution using while to @matle 's solution:

list = [4,4,4,3,3]
condition = 11
def for_fun(list, condition):
    sum_of_sizes = 0
    index = 0
    while True:
        if sum_of_sizes + list[index] > condition:
            break
        sum_of_sizes += list[index]
        index += 1
    return sum_of_sizes
        
print(for_fun(list, condition))

And please try to avoid overwriting python keywords like list !