algorithm-class/体系学习班/class26/Code02_FibonacciProblem.java

package class26;

public class Code02_FibonacciProblem {

	public static int f1(int n) {
		if (n < 1) {
			return 0;
		}
		if (n == 1 || n == 2) {
			return 1;
		}
		return f1(n - 1) + f1(n - 2);
	}

	public static int f2(int n) {
		if (n < 1) {
			return 0;
		}
		if (n == 1 || n == 2) {
			return 1;
		}
		int res = 1;
		int pre = 1;
		int tmp = 0;
		for (int i = 3; i <= n; i++) {
			tmp = res;
			res = res + pre;
			pre = tmp;
		}
		return res;
	}

	// O(logN)
	public static int f3(int n) {
		if (n < 1) {
			return 0;
		}
		if (n == 1 || n == 2) {
			return 1;
		}
		// [ 1 ,1 ]
		// [ 1, 0 ]
		int[][] base = { 
				{ 1, 1 }, 
				{ 1, 0 } 
				};
		int[][] res = matrixPower(base, n - 2);
		return res[0][0] + res[1][0];
	}

	public static int[][] matrixPower(int[][] m, int p) {
		int[][] res = new int[m.length][m[0].length];
		for (int i = 0; i < res.length; i++) {
			res[i][i] = 1;
		}
		// res = 矩阵中的1
		int[][] t = m;// 矩阵1次方
		for (; p != 0; p >>= 1) {
			if ((p & 1) != 0) {
				res = product(res, t);
			}
			t = product(t, t);
		}
		return res;
	}

	// 两个矩阵乘完之后的结果返回
	public static int[][] product(int[][] a, int[][] b) {
		int n = a.length;
		int m = b[0].length;
		int k = a[0].length; // a的列数同时也是b的行数
		int[][] ans = new int[n][m];
		for(int i = 0 ; i < n; i++) {
			for(int j = 0 ; j < m;j++) {
				for(int c = 0; c < k; c++) {
					ans[i][j] += a[i][c] * b[c][j];
				}
			}
		}
		return ans;
	}

	public static int s1(int n) {
		if (n < 1) {
			return 0;
		}
		if (n == 1 || n == 2) {
			return n;
		}
		return s1(n - 1) + s1(n - 2);
	}

	public static int s2(int n) {
		if (n < 1) {
			return 0;
		}
		if (n == 1 || n == 2) {
			return n;
		}
		int res = 2;
		int pre = 1;
		int tmp = 0;
		for (int i = 3; i <= n; i++) {
			tmp = res;
			res = res + pre;
			pre = tmp;
		}
		return res;
	}

	public static int s3(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];
	}

	public static int c1(int n) {
		if (n < 1) {
			return 0;
		}
		if (n == 1 || n == 2 || n == 3) {
			return n;
		}
		return c1(n - 1) + c1(n - 3);
	}

	public static int c2(int n) {
		if (n < 1) {
			return 0;
		}
		if (n == 1 || n == 2 || n == 3) {
			return n;
		}
		int res = 3;
		int pre = 2;
		int prepre = 1;
		int tmp1 = 0;
		int tmp2 = 0;
		for (int i = 4; i <= n; i++) {
			tmp1 = res;
			tmp2 = pre;
			res = res + prepre;
			pre = tmp1;
			prepre = tmp2;
		}
		return res;
	}

	public static int c3(int n) {
		if (n < 1) {
			return 0;
		}
		if (n == 1 || n == 2 || n == 3) {
			return n;
		}
		int[][] base = { 
				{ 1, 1, 0 }, 
				{ 0, 0, 1 }, 
				{ 1, 0, 0 } };
		int[][] res = matrixPower(base, n - 3);
		return 3 * res[0][0] + 2 * res[1][0] + res[2][0];
	}

	public static void main(String[] args) {
		int n = 19;
		System.out.println(f1(n));
		System.out.println(f2(n));
		System.out.println(f3(n));
		System.out.println("===");

		System.out.println(s1(n));
		System.out.println(s2(n));
		System.out.println(s3(n));
		System.out.println("===");

		System.out.println(c1(n));
		System.out.println(c2(n));
		System.out.println(c3(n));
		System.out.println("===");

	}

}