如何找到总和最接近给定数字的 3 个数字

Question

我正在尝试为该问题编写简单的代码。如果我得到一个数组和数字，我需要找到它们的总和接近给定数字的 3 个数字。

我想过先弹出最后一位（第一个数字）然后我有一个没有这个数字的新数组。所以现在我寻找第二个需要小于总和目标的数字。所以我只取比第二个=总和第一个数字小的小数字（但我不知道如何选择它。

最后一个数字会是third=sum-first-second

我尝试编写代码，但它不起作用，而且它非常基础

def f(s,target):
s=sorted(s)
print(s)
print(s[0])
closest=s[0]+s[1]+s[2]
m=s[:-1]
print(m)
for i in range(len(s)):
    for j in range(len(m)):
        if (closest<=target-m[0]) and s[-1] + m[j] == target:
    print (m[j])

n = m[:j] + nums[j+1:]
for z in range (len(z)):
    if (closest<target-n[z]) and s[-1]+ m[j]+n[z] == target:
    print (n[z])



 
s=[4,2,12,3,4,8,14]
target=20
f(s,target)

如果您知道要在此处更改什么。请告诉我谢谢

Answer 1

是这样的吗？

import math

num_find = 1448
lst_Results = []
i_Number = num_find

while i_Number > 0:
    num_Exp = math.floor(math.log(i_Number) / math.log(2))
    lst_Results.append(dict({num_Exp: int(math.pow(2, num_Exp))}))
    i_Number = i_Number - math.pow(2, num_Exp)

print(lst_Results)

在数字序列中：例如 1、2、4、8、16、32、64、128、256、512、1024、2048 等...

前一个数字的总和永远不会大于下一个。这给了我们组合的可能性，例如：数字：1448，除了前面数字的和：8 + 32 + 128 + 256 + 1024

，没有其他组合

然后你找到和提供的数字相近的数字

Answer 2

使用 itertools.combinations to get all combinations of your numbers without replacement of a certain length (three in your case). Then take the three-tuple for which the absolute value of the difference of the sum and target is minimal. min 可以使用 key 参数来指定传递给函数的 iterable 的顺序。

from typing import Sequence, Tuple

def closest_to(seq: Sequence[float], target: float, length: int = 3) -> Tuple[float]:
    from itertools import combinations

    combs = combinations(seq, length)
    diff = lambda x: abs(sum(x) - target)

    return min(combs, key=diff)

closest_to([4,2,12,3,4,8,14], 20) # (4, 2, 14)

这不是最快或最有效的方法，但它在概念上简单而简短。

Answer 3

这是我的解决方案，我试图最大限度地提高代码的性能以不重复任何组合。如果您有任何问题，请告诉我。祝你好运。

def find_3(s,target):

 to_not_rep=[] #This list will store all combinations without repetation
 close_to_0=abs(target - s[0]+s[1]+s[2]) #initile 
 There_is_one=False #False: don't have a combination equal to the target yet
 for s1,first_n in enumerate(s):
    for s2,second_n in enumerate(s):
        if (s1==s2) : continue #to not take the same index
        for s3,third_n in enumerate(s):
            if (s1==s3) or (s2==s3) : continue #to not take the same index
            val=sorted([first_n,second_n,third_n]) #sorting  
            if val in to_not_rep :continue #to not repeat the same combination with diffrent positions  
            to_not_rep.append(val)#adding all the combinations without repetation
            sum_=sum(val) #the sum of the three numbers
            # Good one
            if sum_==target:
                print(f"Found a possibility: {val[0]} + {val[1]} + {val[2]} = {target}")
                There_is_one = True 
            
            if There_is_one is False: #No need if we found combination equal to the target
                # close to the target
                # We know that (target - sum) should equal to 0 otherwise :
                # We are looking for the sum of closet combinations(in abs value) to 0
                pos_n=abs(target-sum_)
                if pos_n < close_to_0:
                    closet_one=f"The closet combination to the target is: {val[0]} + {val[1]} + {val[2]} = {sum_} almost {target} "
                    close_to_0=pos_n
 # Print the closet combination to the target in case we did not find a combination equal to the target 
 if There_is_one is False: print(closet_one)

所以我们可以测试它:

s =[4,2,3,8,6,4,12,16,30,20,5]
target=20
find_3(s,target)
#Found a possibility: 4 + 4 + 12 = 20
#Found a possibility: 2 + 6 + 12 = 20
#Found a possibility: 3 + 5 + 12 = 20

另一个测试：

s =[4,2,3,8,6,4,323,23,44]
find_3(s,target)
#The closet combination to the target is: 4 + 6 + 8 = 18 almost 20

Answer 4

这是一个returns所有可能性的简单解决方案。您的案例在 0.002019 秒内完成

from itertools import combinations
import numpy as np
def f(s, target):
    dic = {}
    for tup in combinations(s, 3):
        try:
            dic[np.absolute(np.sum(tup) - target)].append(str(tup))
        except KeyError:
            dic[np.absolute(np.sum(tup) - target)] = [tup]
    print(dic[min(dic.keys())])

如何找到总和最接近给定数字的 3 个数字

how to find 3 Numbers with Sum closest to a given number

python

arrays

python-2.7

python-3.x

pandas