题解 | #0左边必有1的二进制字符串的数量#

/*方法3：快速幂
    时间复杂度O(logN)
    取模的话在乘法除加上即可
    * */
    public static int getNum3(int n) {
        if (n < 1) {
            return 0;
        }
        if (n == 1 || n == 2) {
            return n;
        }
        int[][] base = {{1, 1}, {1, 0}};
        int[][] res = matrixPower(base, n - 2);
        return 2 * res[0][0] + res[1][0];
    }

    //求矩阵m的p次方
    public static int[][] matrixPower(int[][] m, int p) {
        int[][] res = new int[m.length][m[0].length];
        //先把res设为单位矩阵，相当于整数中1
        for (int i = 0; i < m.length; i++) {
            res[i][i] = 1;
        }

        int[][] tmp = m;
        //求p对应的二进制
        for (; p != 0; p >>= 1) {
            if ((p & 1) != 0) {
                res = mulMatrix(res, tmp);
            }
            tmp = mulMatrix(tmp, tmp);
        }

        return res;
    }

    //矩阵m1 矩阵m2相乘
    public static int[][] mulMatrix(int[][] m1, int[][] m2) {
        int[][] res = new int[m1.length][m2[0].length];
        for (int i = 0; i < m1.length; i++) {
            for (int j = 0; j < m2[0].length; j++) {
                for (int k = 0; k < m2.length; k++) {
                    res[i][j] = res[i][j] + m1[i][k] * m2[k][j];
                }
            }
        }

        return res;
    }