1652. Defuse the Bomb

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

这题只要算出对应的索引范围即可。

class DefuseTheBomb : public Solution {
public:
    void Exec() {

    }
    vector<int> decrypt(vector<int>& code, int k) {
        vector<int> res(code.size(), 0);
        for (int i = 0; i < code.size(); i++) {
            int sum = 0;
            if (k != 0) {
                for (int j = 0; j < abs(k); j++) {
                    int index = i + (k > 0 ? j + 1 : -j - 1);
                    if (index < 0) {
                        index += code.size();
                    }
                    index = index % code.size();
                    sum += code[index];
                }
                res[i] = sum;
            }
        }
        return res;
    }
};