返回執行 k 次逆時針迴圈輪轉操作後的矩陣。
2 <= m, n <= 50，1 <= k <= 109

思路：每一環的寬：n-2i，2i是因為，每次要在左、右邊都留出i個位置，所以乘以2

class Solution {
public:
    vector<int> change(vector<int> a, int k) {
        deque<int> q(a.begin(), a.end());
        k %= a.size();
        while (k--) {
            int v = q.front(); q.pop_front();
            q.push_back(v);
        }
        vector<int> ans(q.begin(), q.end());
        return ans;
    }
    vector<vector<int>> rotateGrid(vector<vector<int>>& g, int k) {
        int n = g.size(), m = g[0].size();
        int round = min(n, m) / 2;
        int i = 0, j = 0;

        while (round--) {
            int idx = 0;
            vector<int> data((n - 2*i) * (m - 2*j) - (n - 2*(i+1)) * (m - 2*(j+1)));
            for (int y = j; y < m - j - 1; y++) {
                data[idx++] = g[i][y];
            }
            for (int x = i; x < n - i - 1; x++) {
                data[idx++] = g[x][m-j-1];
            }
            for (int y = m - j - 1; y > j; y--) {
                data[idx++] = g[n-i-1][y];
            }
            for (int x = n - i - 1; x > i; x--) {
                data[idx++] = g[x][j];
            }
            auto after = change(data, k);
            idx = 0;
            for (int y = j; y < m - j - 1; y++) {
                g[i][y] = after[idx++];
            }
            for (int x = i; x < n - i - 1; x++) {
                g[x][m-j-1] = after[idx++];
            }
            for (int y = m - j - 1; y > j; y--) {
                g[n-i-1][y] = after[idx++];
            }
            for (int x = n - i - 1; x > i; x--) {
                g[x][j] = after[idx++];
            }
            i++, j++;
        }
        return g;
    }
};