如何对单个列表中的两个元素求和并与 integer 参数进行比较

Question

I'm writing a function that checks to see if the sum of any two elements within the same list equals an integer argument.

For example: Given 17 and [10, 15, 3, 7], it should return True as 17 is 10 + 7.

Given 17 and [10, 15, 4, 8], it should return False, as there is no pair in the list that sums to 17.

I'm completely lost on how to approach this. Any help is appreciated!

for i in range(0, len(lst) - 1):
    return k == lst[i] + lst[i]

return lst

Given 17 and [10, 15, 3, 7], it should return True as 17 is 10 + 7.

Given 17 and [10, 15, 4, 8], it should return False, as there is no pair in the list that sums to 17.

Answer 1

def check(n, lst):
    lst_set = set(lst)
    for i in lst:
        if (n-i) in lst_set: 
            if (n-i) != i:   # check to see if it returned itself (not a pair)
                return True
    return False

Here you have n as the goal number and then lst as your input.

You convert the input to a set first to speed up lookups, then for each one you take your goal number, subtract your current number, and see if it's in the set. If it is, then you have an answer True and if not, move on. If you reach the end without finding one then you don't and return False

The main idea behind this is that when you take your goal number and subtract the number you are on now, then you get the number you need to make a pair. So you check your set to see if that number exists in it. You don't need to go through each potential pair this way.

---

Edit: If you want to handle duplicate inputs then you can build the set as you go which will be slower than the above answer which cannot handle duplicate inputs but will still be extremely fast compared to non-hash table implementations.

def check_withdup(n, lst):
    lst_set = set()   # start with blank set
    for i in lst:
        if (n-i) in lst_set:   # if it is in there, it wasn't itself so it is a pair
            return True
        else:
            lst_set.add(i)   # if it isn't in there, add it in
    return False

This will work in one iteration but the time to set.add() is slower than the time to create a set from the same list which is why the above will be faster than this.

Answer 2

def check(n, lst):

    for i in lst:
        temp = lst.copy()
        temp.remove(i)
        if n-i in temp:
            return True
    return False


l = [10,6,7,5,10]

print(check(20,l))

this returns

True

Answer 3

def checkSum(num, lst):
    for i in lst:
       for j in lst:
         if num==i + j:
            return True

return False

This way is not as efficient as the previous answer, but it is more a Brute Force approach comparing sum of all the pairs and just checking if they satisfy the condition.

Answer 4

Try this bro

 def check(n,lst):
        for item in lst:
            temp = item
            for i in range(len(lst)):
                if lst.index(temp) != i: #This is to avoid comparison with the same index
                    if n == temp + lst[i]:
                        print("MATCH FOUND! ", temp, " and ", lst[i])
                        return True
                    else:
                        continue