leetcode-1629-easy

Slowest Key

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

思路一：复制两遍数组，根据情况遍历即可

    public int[] decrypt(int[] code, int k) {
        int[] array = new int[code.length * 2];
        System.arraycopy(code, 0, array, 0, code.length);
        System.arraycopy(code, 0, array, code.length, code.length);

        if (k == 0) {
            Arrays.fill(code, 0);
        } else if (k > 0) {
            for (int i = 0; i < code.length; i++) {
                int sum = 0;
                for (int j = 1; j <= k; j++) {
                    sum += array[i + j];
                }
                code[i] = sum;
            }
        } else {
            for (int i = code.length; i < array.length; i++) {
                int sum = 0;

                for (int j = 1; j <= Math.abs(k); j++) {
                    sum += array[i - j];
                }

                code[i - code.length] = sum;
            }
        }

        return code;
    }