给定一个数组，找到总和等于给定总和的元素对，并返回其索引的总和

Question

大家在问题中读到的大家好，我试图在数组中找到等于给定总和的元素对，并返回它们各自索引的总和。

我能够返回给定总和的元素对，但未能返回其索引的总和。 这是我的代码：

arr = [1, 4, 2, 3, 0 , 5]
sum = 7

x = min(arr)
y = max(arr)

while x < y:
    if x + y > sum:
        y -= 1
    elif x + y < sum:
        x += 1
    else:
        print("(", x, y, ")")
        x += 1

我的输出：

( 2 5 )    
( 3 4 )

这是我需要做的进一步工作：

2 + 5 = 7→索引2 + 5 = 7;

3 + 4 = 7→索引3 +1 = 4;

7 + 4 = 11→返回11;

提前致谢！

Answer 1

您可以尝试使用嵌套循环：

arr = [1, 4, 2, 3, 0 , 5]
sums = 7
tlist = []
for i in range(len(arr)):
    for j in range(len(arr)-1):
        if (i!=j) and ((arr[i] + arr[j+1]) == sums):
            if (i,j+1) not in tlist and (j+1,i) not in tlist:
                tlist.append((i,j+1))
                print("index ->",i,"   ",j+1)
                print("sum=", i+j+1)

输出：

index -> 1     3
sum= 4
index -> 2     5
sum= 7

Answer 2

您可以使用itertools轻松检查combinations sum ，例如，

>>> import itertools
>>> num = 7
>>> for a,b in itertools.combinations(arr, 2):
...   if a + b == num:
        aindex, bindex = arr.index(a), arr.index(b)
...     indices_sum = aindex + bindex
...     print('[element sum]: {} + {} = {} [indices sum]: {} + {} = {}'.format(a, b, a + b, aindex, bindex , indices_sum))
... 
[element sum]: 4 + 3 = 7 [indices sum]: 1 + 3 = 4
[element sum]: 2 + 5 = 7 [indices sum]: 2 + 5 = 7
>>> arr
[1, 4, 2, 3, 0, 5]

Answer 3

您可以通过计算差异然后检查第一个数组中是否存在每个元素来采用不同的方法。

arr = [1, 4, 2, 3, 0, 5]
the_sum = 7

diff = [the_sum - x for x in arr]
for idx, elem in enumerate(diff):
    try:
        index = arr.index(elem)
        sum_of_indices = idx + index
        print("{} + {} = {}".format(idx, index, sum_of_indices))
    except ValueError:
        pass

输出

1 + 3 = 4
2 + 5 = 7
3 + 1 = 4
5 + 2 = 7

要删除重复项，总是很容易对索引元组进行frozenset设置

a = [(2,1), (1,2), (3,2), (2,3)]
{frozenset(x) for x in a} # {frozenset({2, 3}), frozenset({1, 2})}