描述
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.
Note:
- n == code.length
- 1 <= n <= 100
- 1 <= code[i] <= 100
- -(n - 1) <= k <= n - 1
解析
根据题意，将两个 code 拼起来看成首尾相连的环，如果 k > 0 ，则索引为 i 元素的值解码为 sum(code[i+1:i+1+k]) ，如果 k < 0 ，则索引为 i 元素解码为 sum(code[i+k:i]) ，如果 k == 0 ，则返回同样长度的都为 0 的数组，，最后返回解码后的数组，。
解答
class Solution(object):
def decrypt(self, code, k):
"""
:type code: List[int]
:type k: int
:rtype: List[int]
"""
n = len(code)
if k==0:
return [0] * n
result = []
code += code
if k>0:
for i in range(n):
result.append(sum(code[i+1:i+1+k]))
else :
for i in range(n, 2*n):
result.append(sum(code[i+k:i]))
return result
运行结果
Runtime: 52 ms, faster than 15.58% of Python online submissions for Defuse the Bomb.
Memory Usage: 13.5 MB, less than 48.48% of Python online submissions for Defuse the Bomb.
原题链接：https://leetcode.com/problems/defuse-the-bomb