題目如下：

You have a bomb to defuse, and your time is running out! Your informer will provide you with acirculararraycodeof length ofnand a keyk.

To decrypt the code, you must replace every number. All the numbers are replacedsimultaneously.

• Ifk > 0, replace theithnumber with the sum of thenextknumbers.
• Ifk < 0, replace theithnumber with the sum of thepreviousknumbers.
• Ifk == 0, replace theithnumber with0.

Ascodeis circular, the next element ofcode[n-1]iscode[0], and the previous element ofcode[0]iscode[n-1].

Given thecirculararraycodeand an integer keyk, returnthe decrypted code to defuse the bomb

!

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.

Constraints:

• n == code.length
• 1 <= n<= 100
• 1 <= code[i] <= 100
• -(n - 1) <= k <= n - 1

解題思路：很簡單的題目。

程式碼如下：

class Solution(object):
    def decrypt(self, code, k):
        """
        :type code: List[int]
        :type k: int
        :rtype: List[int]
        """
        if k == 0:return [0] * len(code)
        ori_len = len(code)
        code_refactor = code * 2

        if k > 0:
            for i in range(ori_len):
                code[i] = sum(code_refactor[i+1:i+1+k])
            return code
        else:
            k = -k
            for i in range(ori_len,len(code_refactor)):
                code[i - ori_len] = sum(code_refactor[i-k:i])
            return code