Sum of Pairs
Codewars Kata 32√
Description
https://www.codewars.com/kata/54d81488b981293527000c8f
-简述：本题给定一列数字和一个求和的值，返回数列中前两个相加的和为该数字的值。
-思路：已知一个数，判断另一个数是否在数列中
-难点：1 防止超时
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]
sum_pairs([4, 3, 2, 3, 4], 6) 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) Entire pair is earlier, and therefore is the correct answer == [3, 7] Negative numbers and duplicate numbers can and will appear.
NOTE: There will also be lists tested of lengths upwards of 10,000,000 elements. Be sure your code doesn’t time out.
My solution (error)
# 错误回答，超时了
def sum_pairs(ints,s):
for i in range(0,len(ints)):
if s-ints[i] in ints[i+1:]:
return [ints[i],s-ints[i]]
Given solution
def sum_pairs(ints, s):
rst = set()
for i in range(0,len(ints)):
if s - ints[i] in rst:
return [s - ints[i],ints[i]]
rst.add(ints[i])
set是一个无序且不重复的元素集合。 注意在创建空集合的时候只能使用s=set()，因为s={}创建的是空字典
Points
1 set()的时间复杂度是O(1)
PS: 今天我的方法timed out了，看了答案才做出来。
