斐波那契系列问题的递归和动态规划

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//斐波那契系列问题的递归和动态规划public class FibonacciProblem{//************************************************************    //递归解法O(2的n次方)    public static int f1(int n)    {    if(n<1)    {    return 0;    }    if(n==1||n==2)    {    return 1;    }    return f1(n-2)+f1(n-1);    }    //迭代解法(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 tmp=0;    for(int i=3;i<=n;i++)    {    tmp=res;    res=res+pre;    pre=tmp;    }    return res;    }   //矩阵乘法(O(log(N)))    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];// 先把res设为单位矩阵，相等于整数中的1。for (int i = 0; i < res.length; i++) {res[i][i] = 1;}int[][] tmp = m;for (; p != 0; p >>= 1) {if ((p & 1) != 0) {res = muliMatrix(res, tmp);}tmp = muliMatrix(tmp, tmp);}return res;}   //矩阵的乘法public static int[][] muliMatrix(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] += m1[i][k] * m2[k][j];}}}return res;}//*****************************************************************    //上台阶问题    //递归解法    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);}   //迭代解法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;}   //矩阵的乘法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];}//****************************************************************//生奶牛问题//递归解法    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);}   //迭代解法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;}   //三阶递推式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){//System.out.println("Hello"); int n = 30;//经典fibonacciSystem.out.println(f1(n));System.out.println(f2(n));System.out.println(f3(n));System.out.println("===");                //上台阶问题System.out.println(s1(n));System.out.println(s2(n));System.out.println(s3(n));System.out.println("===");                //生奶牛问题System.out.println(c1(n));System.out.println(c2(n));System.out.println(c3(n));System.out.println("===");}}
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