Fibonacci数列及相关的问题的描述#

给定正整数n, 求解Fibonacci数列第n项的值;
给定正整数N, 代表台阶, 一次可以跨2个或者1个台阶, 有多少走法
假设成熟的母牛只会生1头小母牛, 并且永远不会死, 第一年农场有1只成熟的母牛,从第二年开始, 母牛开始生小母牛. 每只小母牛3年之后成熟. 给定正整数N, 求出N年后牛的数量.

基础解法#

递归的解法#

递归的解法是最基础, 也是最好理解, 但是时间复杂度很高, 为 $O(2^n)$ ,空间复杂度为 $O(2^n)$

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

迭代的解法#

迭代的解法即将已经解决的位置处的数存储, 之后直接调用, 不用再计算

时间复杂度为 $O(n)$ , 空间复杂度为 $O(n)$ .

public static int fibIteration(int n) {
    if (n == 1) {
        return 1;
    }
    if (n == 0) {
        return 0;
    }
    int[] arr = new int[n + 1];
    arr[0] = 0;
    arr[1] = 1;
    for (int i = 2; i < n + 1; i++) {
        arr[i] = arr[i - 1] + arr[i - 2];
    }
    return arr[n];
}

仅记录前两个值, 空间复杂度为 $O(1)$

public static BigInteger fibIteration(int n) {
        if (n == 1) {
            return new BigInteger("1");
        }
        if (n == 0) {
            return new BigInteger("1");
        }

        BigInteger res = new BigInteger("0");
        BigInteger pre_1 = BigInteger.valueOf(1);
        BigInteger pre_2 = BigInteger.valueOf(0);

        for (int i = 2; i < n + 1; i++) {
            res = pre_1.add(pre_2);
            pre_2 = pre_1;
            pre_1 = res;
        }
        return res;
    }

进阶解法#

推导过程#

$F(n)=F(n-1)+F(n-2) \Rightarrow \begin{vmatrix} F(n) & F(n-1) \end{vmatrix}#

\begin{vmatrix} F(n) & F(n-1) \end{vmatrix} \times \begin{vmatrix} a & b \ c & d \end{vmatrix}$

$F(n)=F(n-1)+F(n-2) \\\Rightarrow\begin{vmatrix}F(n) & F(n-1) \\\end{vmatrix}=\begin{vmatrix}F(n) & F(n-1) \end{vmatrix}\times\begin{vmatrix}a & b \\c & d \end{vmatrix}\\$

解得:

$a=1 \\ b=1 \\ c=1 \\ d=0$

即:

$\begin{vmatrix} F(n) & F(n-1)\end{vmatrix}=\begin{vmatrix}F(n-1)& F(n-2)\end{vmatrix}\times\begin{vmatrix}1 & 1 \\\\ 1 & 0\end{vmatrix}\\\\\Rightarrow\begin{vmatrix} F(n) & F(n-1)\end{vmatrix}=\begin{vmatrix} 1 & 1\end{vmatrix}\times{\begin{vmatrix}1 & 1 \\\\ 1 & 0\end{vmatrix}}^{n-2}$

代码#


public static BigInteger fibMatrix(int n) {
    if (BigInteger.ZERO.compareTo(BigInteger.valueOf(n)) == 0) {
        return new BigInteger("0");
    }
    if (BigInteger.ONE.compareTo(BigInteger.valueOf(n)) == 0) {
        return new BigInteger("1");
    }
    return matrixPower(new int[][]{{1, 1}, {1, 0}}, n)[0][0];
}

//计算矩阵matrix的p次方
public static BigInteger[][] matrixPower(int[][] matrix, int p) throws IllegalArgumentException {
    if (matrix[0].length != matrix.length) {
        throw new IllegalArgumentException("矩阵输入错误, 无法进行乘法.");
    }
    BigInteger[][] res = new BigInteger[matrix.length][matrix.length];
    for (int i = 0; i < res.length; i++) {
        for (int j = 0; j < res[0].length; j++) {
            res[i][j] = BigInteger.valueOf(0);
        }
    }
    for (int i = 0; i < res.length; i++) {
        res[i][i] = new BigInteger("1");
    }

    BigInteger[][] tmp = new BigInteger[matrix.length][matrix.length];
    for (int i = 0; i < tmp.length; i++) {
        for (int j = 0; j < tmp[0].length; j++) {
            tmp[i][j] = BigInteger.valueOf(matrix[i][j]);
        }
    }

    while (p > 0) {
        int flag = p & 1;
        if (flag == 1) {
            res = multiMatrix(res, tmp);
        }
        tmp = multiMatrix(tmp, tmp);
        p = p >> 1;
    }

    return multiMatrix(new BigInteger[][]{{new BigInteger("1"), new BigInteger("1")}}, res);
}

//矩阵乘法
public static BigInteger[][] multiMatrix(BigInteger[][] m1, BigInteger[][] m2) throws IllegalArgumentException {
    if (m1[0].length != m2.length) {
        throw new IllegalArgumentException("矩阵输入错误, 无法进行乘法.");
    }
    BigInteger[][] res = new BigInteger[m1.length][m2[0].length];
    for (int i = 0; i < m1.length; i++) {
        for (int j = 0; j < m2[0].length; j++) {
            res[i][j] = new BigInteger("0");
        }
    }

    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].add(m1[i][k].multiply(m2[k][j]));
            }
        }
    }
    return res;
}

运行时间比较:#

当n=45时, 递归的方法就需要6794ms才能完成.

当n=1000000时, 迭代的方法耗时16s左右, 使用矩阵的方法只需耗时1.266s即完成