给定一个整数数组，找到两个差最小且和为M的整数

Question

给定一个无序整数列表，以这种方式打印两个总计为m的整数int1 int2（int 1小于或等于int 2，它们之间用空格隔开）。假设整数列表中总是有m的解。int1和int2必须是正整数。

如果有多个解决方案，请打印差异最小的整数对。

例子：

Li = [2, 6, 8, 10, 4]
m = 10

Output: 
4 6

这是我的代码，但是根据我们的编译器（这是我们班的一个练习），我对隐藏的测试用例有错误的输出。

li = [2, 6, 8, 10, 4]
m = int(input())
selection = {}                  # I created a dictionary to put all pairs, sort them
                                # and print the pair with the least difference.

for j in li:

index = li.index(j)             
temp = li.pop(index)

    if m-j in li and j != m:          # checks if m-j is present in the list, and j
                                      # does not equal to m, since we're only looking
                                      # for positive integer pairs

        if abs(j - (m-j)) not in selection :            # don't have to rewrite in the dict
                                                             # if pair is already written.

            selection[abs(j - (m-j))] = [min(j, m-j), max(j, m-j)]

    li.insert(index, temp)

output = min(selection.keys()) # gets the pair  with the minimum difference in the dict

print(*selection[output])      

更新：代码工作！

Answer 1

代码：

Li = [2, 6, 8, 10, 4]
m = 10

#Output: 
#4 6

m_satisfied = []
for i in range(len(Li) - 1):
    for j in range(i+1, len(Li)):
        if (Li[i] + Li[j]) == m:
            m_satisfied.append(tuple([Li[i], Li[j]]))

diff_dict = {i: abs(t[0] - t[1]) for i,t in enumerate(m_satisfied)}
least_diff = min(diff_dict.values())
ind_m_satisfied = [k for k, v in diff_dict.items() if v == least_diff]
res_int = [m_satisfied[i] for i in ind_m_satisfied]
print(res_int)
print(res_int[0][0], res_int[0][1])

输出：

[(6, 4)]
6 4