시간 보정

코드 출현 2018 1일 차

작업: X에 대해 풀기 여기서...

1 부

X = the resulting frequency - starting from 0 - after all of the changes in frequency have been applied

2 부

X = the first frequency my device reaches twice

예시 입력

+1
-2
+3
+1

1 부

세 친구: split() , map() , reduce()

Generate an ordered list of numbers from the input
For each number
  Accumulate a sum - starting at 0
Return the sum

input.split('\n').map(Number).reduce((a,c) => a + c)

2 부

돌고 돌고, 겉보기에는 영원할 것 같아

Set a list of frequency history, starting with one element: 0
Set frequency as 0
Set index as 0
Do as long as the last item in the list is not found elsewhere in the list
  If index is equal to the length of the list
    Reset index to 0
  Increment frequency by the number change at the current index in the list of frequency changes
  Insert the new frequency at the end of history
  Increment index by 1
Return the last item in the history list

해냈어!!

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