如何找到加起来等于给定总和的组合和相应的索引?而且,是否可以处理大小为500000(较大大小)的元素的列表?
输入:
l1 = [9,1, 2, 7, 6, 1, 5]
target = 8
**Constraints**
1<=(len(l1))<=500000
1<=each_list_element<=1000
输出:
Format : {index:element}
{1:1, 5:1, 4:6} #Indices : 1,5,4 Elements : 1,1,6
{1:1, 2:2, 6:5}
{5:1, 2:2, 6:5}
{1:1, 3:7}
{5:1, 3:7}
{2:2, 4:6}
已尝试:
from itertools import combinations
def test(l1, target):
l2 = []
l3 = []
if len(l1) > 0:
for r in range(0,len(l1)+1):
l2 += list(combinations(l1, r))
for i in l2:
if sum(i) == target:
l3.append(i)
return l3
l1 = [9,1, 2, 7, 6, 1, 5]
target = 8
print(test(l1,target))
[(1, 7), (2, 6), (7, 1), (1, 2, 5), (1, 6, 1), (2, 1, 5)]
有人可以引导我吗?
UPDATE
Apart from above, code fails to handle these scenarios
Input = [4,6,8,5,3]
target = 3
Outputs {} , need to output {4:3}
Input = [4,6,8,3,5,3]
target = 3
Outputs {} , need to output {5:3,3:3} #corrected index
Input = [1,2,3,15]
target = 15
Outputs = {}, need to output {3:15}
您的代码很接近,我将使用枚举以元组对的形式获取索引和值。我总是删除任何索引和值元组,其中该值大于目标值,因为这不可能匹配。这将产生较少的组合。然后像您一样,我只遍历元组的排列并对每个排列中的值求和,如果它求和到目标,则产生该排列。最后在循环中输出值,我将烫发给dict转换为想要的dict格式
from itertools import combinations
def find_sum_with_index(l1, target):
index_vals = [iv for iv in enumerate(l1) if iv[1] < target]
for r in range(1, len(index_vals) + 1):
for perm in combinations(index_vals, r):
if sum([p[1] for p in perm]) == target:
yield perm
l1 = [9, 1, 2, 7, 6, 1, 5]
target = 8
for match in find_sum_with_index(l1, target):
print(dict(match))
输出
{1: 1, 3: 7}
{2: 2, 4: 6}
{3: 7, 5: 1}
{1: 1, 2: 2, 6: 5}
{1: 1, 4: 6, 5: 1}
{2: 2, 5: 1, 6: 5}
您可以仅使用索引函数来获取索引,并在字典的帮助下将它们存储为键:值对,位于另一个列表中,如下所示,
from itertools import combinations
def test(l1, target):
l2 = []
l3 = []
l4=[]
dict1={}
a=0
if len(l1) > 0:
for r in range(0,len(l1)+1):
l2 += list(combinations(l1, r))
for i in l2:
dict1={}
if sum(i) == target:
for j in i:
a=l1.index(j)
dict1[a]=j
l4.append(dict1)
l3.append(i)
return l4
l1 = [4,6,8,5,3]
target = 3
print(test(l1,target))
输出:
[{4: 3}]
如您所见,l1 = [4,6,8,5,3] target = 3
的条件以前不起作用。
希望这会有所帮助!