给定一个数组，找到总和等于给定总和的元素对，并返回它们的索引总和

Question

大家好，正如您在问题中所读到的，我正在尝试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，但未能返回它们索引的总和。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 )

这是我需要进一步做的：

2 + 5 = 7 → 指数 2 + 5 = 7;

3 + 4 = 7 → 指数 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sum 中的combinations 之类的，

>>> 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每个元素是否存在于第一个数组中来采取不同的方法。

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，获取索引元组的frozenset 总是很容易

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