1652. Defuse the Bomb

# 1652. Defuse the Bomb#

## 题目 #

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`

## 题目大意 #

• 如果 k > 0 ，将第 i 个数字用 接下来 k 个数字之和替换。
• 如果 k < 0 ，将第 i 个数字用 之前 k 个数字之和替换。
• 如果 k == 0 ，将第 i 个数字用 0 替换。

## 解题思路 #

• 给出一个 code 数组，要求按照规则替换每个字母。
• 简单题，按照题意描述循环即可。

## 代码 #

``````package leetcode

func decrypt(code []int, k int) []int {
if k == 0 {
for i := 0; i < len(code); i++ {
code[i] = 0
}
return code
}
count, sum, res := k, 0, make([]int, len(code))
if k > 0 {
for i := 0; i < len(code); i++ {
for j := i + 1; j < len(code); j++ {
if count == 0 {
break
}
sum += code[j]
count--
}
if count > 0 {
for j := 0; j < len(code); j++ {
if count == 0 {
break
}
sum += code[j]
count--
}
}
res[i] = sum
sum, count = 0, k
}
}
if k < 0 {
for i := 0; i < len(code); i++ {
for j := i - 1; j >= 0; j-- {
if count == 0 {
break
}
sum += code[j]
count++
}
if count < 0 {
for j := len(code) - 1; j >= 0; j-- {
if count == 0 {
break
}
sum += code[j]
count++
}
}
res[i] = sum
sum, count = 0, k
}
}
return res
}
``````

Apr 8, 2023