QUESTION:
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
EXPLANATION:
这道题目的难点其实就在于数组的长度. 那么如果k大于0超过了数组长度, 那么对应的index就是 (index+k) % size. 那么如果k小于0, 到了数组外, 那么对应的index就是 size + (index + k). 这样, 就能够算出对应的和了. 最后再填到结果中即可.
SOLUTION:
class Solution {
fun decrypt(code: IntArray, k: Int): IntArray {
if (k == 0) {
return IntArray(code.size, init = {
0
})
}
var result = code.copyOf()
for (index in code.indices) {
var tmpK = k
var tmp = 0
if (k > 0) {
while (tmpK > 0) {
tmp += code[(index + tmpK) % code.size]
tmpK--
}
} else {
while (tmpK < 0) {
var tmpIndex = if (index + tmpK >= 0) index+tmpK else code.size + (index + tmpK)
tmp += code[tmpIndex]
tmpK++
}
}
result[index] = tmp
}
return result
}
}