【leetcode】1652. Defuse the Bomb

题目如下：

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!

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

解题思路：很简单的题目。

代码如下：

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

标签：code,Bomb,sum,number,len,numbers,ori,1652,Defuse
来源： https://www.cnblogs.com/seyjs/p/14931283.html