题目
（5kyu）Given a list of integers and a single sum value, return the first two values (parse from the left please) in order of appearance that add up to form the sum.
sum_pairs([11, 3, 7, 5], 10)
# ^--^ 3 + 7 = 10
== [3, 7]
sum_pairs([4, 3, 2, 3, 4], 6)
# ^-----^ 4 + 2 = 6, indices: 0, 2 *
# ^-----^ 3 + 3 = 6, indices: 1, 3
# ^-----^ 2 + 4 = 6, indices: 2, 4
# * entire pair is earlier, and therefore is the correct answer
== [4, 2]
sum_pairs([0, 0, -2, 3], 2)
# there are no pairs of values that can be added to produce 2.
== None/nil/undefined (Based on the language)
sum_pairs([10, 5, 2, 3, 7, 5], 10)
# ^-----------^ 5 + 5 = 10, indices: 1, 5
# ^--^ 3 + 7 = 10, indices: 3, 4 *
# * entire pair is earlier, and therefore is the correct answer
== [3, 7]
Negative numbers and duplicate numbers can and will appear.
这一题跟LeetCode的“两数之和”一样。
我的答案
def sum_pairs(ints, s):
cache = set()
for i in ints:
other = s - i
if other in cache:
return [other, i]
cache.add(i)