You have a bomb to defuse, and your time is running out! Your informer will provide you with a?circular?array?code?of length of?n?and a key?k.

To decrypt the code, you must replace every number. All the numbers are replaced?simultaneously.

• If?k > 0, replace the?ith?number with the sum of the?next?k?numbers.

• If?k < 0, replace the?ith?number with the sum of the?previous?k?numbers.

• If?k == 0, replace the?ith?number with?0.

As?code?is circular, the next element of?code[n-1]?is?code[0], and the previous element of?code[0]?is?code[n-1].

Given the?circular?array?code?and an integer key?k, return?the decrypted code to defuse the bomb!

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Example 1:

Input: code = [5,7,1,4], k = 3

Output: [12,10,16,13]

Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.

Example 2:

Input: code = [1,2,3,4], k = 0

Output: [0,0,0,0]

Explanation: When k is zero, the numbers are replaced by 0.

Example 3:

Input: code = [2,4,9,3], k = -2

Output: [12,5,6,13]

Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.

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Constraints:

• n == code.length

• 1 <= n?<= 100

• 1 <= code[i] <= 100

• -(n - 1) <= k <= n - 1

easy 題目，但是寫了2個(gè)函數(shù)，分別是向前的，和向后的。。

Runtime:?2 ms, faster than?39.02%?of?Java?online submissions for?Defuse the Bomb.

Memory Usage:?42.5 MB, less than?45.90%?of?Java?online submissions for?Defuse the Bomb.