斐波那契数列

public static void main(String[] args) {
// System.out.println(0X0F);
/*
* int i =1; i<<=2; System.out.println(i);
*/

}

// O(2^n)
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);
}

// O(N)
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 temp = 0;
for (int i = 3; i <= n; i++) {
temp = res;
res = res + pre;
pre = temp;
}
return res;
}
/**
* 如果严格遵循 : F(N) = F(N-1) + F(N-2)
* 那么 (F(n),F(n-1)) = (F(N-1),F(N-2))*|a b|
*                                     |c d|
* @param n
* @return
*/
//O(logn)
public static int f3(int n) {
if (n < 1) {
return 0;
}
if (n == 1 || n == 2) {
return 1;
}
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];
// 单位矩阵
for (int i = 0; i < res.length; i++) {
res[i][i] = 1;
}
int[][] temp = m;
// 75的二进制1001011
for (; p != 0; p >>= 1) {
if ((p & 1) != 0) {
res = muliMatrix(res, temp);
}
temp = muliMatrix(temp, temp);
}
return res;
}

// 两矩阵相乘
public static int[][] muliMatrix(int[][] m1, int[][] m2) {
int[][] res = new int[m1.length][m2[0].length];
for (int i = 0; i < m2[0].length; i++) {
for (int j = 0; j < m1.length; j++) {
for (int k = 0; k < m2.length; k++) {
res[i][j] += m1[i][k] * m2[k][j];
}
}
}
return res;
}