给定一个数组，找到元素对，其总和等于给定总和和 return 它们索引的总和

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以及 return 它们各自索引的总和。

我能够 return 给定总和的元素对，但未能 return 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 → Return 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 是否有 combinations，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

您可以采用不同的方法，计算差异，然后检查第一个数组中是否存在每个元素。

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})}

给定一个数组，找到元素对，其总和等于给定总和和 return 它们索引的总和

Given an array find element pairs whose sum is equal to the given sum and return the sum of their indices

python

algorithm

pairwise