斐波那契数列(二)

 题目：

   一个人爬楼梯，一步可以迈一级，二级，三级台阶，如果楼梯有N级，编写程序，输出所有走法。java实现。

不考虑顺序的算法（第几次迈几个台阶）

public class Hello {private static final int N = 3000; // 总共有N个台阶private static final int Tai_One = 1; // 一次跨1步台阶private static final int Tai_Two = 2; // 一次跨2步台阶private static final int Tai_Three = 3; // 一次跨3步台阶// 每一种跨步方法单独跨的最大值private static final int Max_One = N / Tai_One;private static final int Max_Two = N / Tai_Two;private static final int Max_Three = N / Tai_Three;public static void main(String[] args) {Hello.method();}/** */public static void method() {int count = 0;long beginT = System.currentTimeMillis();for (int i = 0; i <= Max_One; i++) {for (int j = 0; j <= Max_Two - i / Tai_Two; j++) {for (int k = 0; k <= Max_Three - i / Tai_Three - j / Tai_Three; k++) {if (i * 1 + j * 2 + k * 3 == N) {count++;System.out.println(i + " " + j + " " + k);}}}}long endT = System.currentTimeMillis();System.out.println("total method:" + count + "---耗时："+ (endT - beginT));}}

考虑迈的顺序(以下代码转自点击打开链接)

考虑顺序的情况一种实现方式是通过“构造树”来实现。另一种实现思路就是典型的斐波那契数列应用。

构造树实现:

/* * To change this template, choose Tools | Templates and open the template in * the editor. */package com.csdn.question;import java.util.ArrayList;import java.util.List;/** * Description: 一个人爬楼梯，一步可以迈一级，二级，三级台阶，如果楼梯有N级，编写程序，输出所有走法。java实现。 Thinking: * 先试探性的走出第一步,在第一步的基础上递归处理剩下的台阶数,直到最后走出一棵树,然后遍历树,打印出路径 *  * @author karl */public class ClimbStairs_ThunderSubject {public static final int step_1 = 1; // 一步一级public static final int step_2 = 2; // 一步二级public static final int step_3 = 3; // 一步三级private static int total = 0;/** * Description: 开始走楼梯,即创建基于根节点root的树 *  * @param root *            根节点 * @param n *            楼梯级数 */public static void climbStaris_CreateTree(Node root, int n) {if (n == 1) { // 若台阶数为1,则只能走一步一级Node step = new Node(step_1);root.children.add(step);} else if (n == 2) { // 若台阶数为2,第一步可以走一级,也可以走两阶,用for循环处理for (int i = 1; i <= 2; i++) {Node step = new Node(i);root.children.add(step);// 走完第一步,剩下的台阶数递归处理int left = n - i;climbStaris_CreateTree(step, left);}} else if (n >= 3) { // 若台阶数大于等于3,则第一步有三种走法,一级,二级,或三级,for循环处理for (int i = 1; i <= 3; i++) {Node step = new Node(i);root.children.add(step);// 走完第一步,剩下的台阶数递归处理int left = n - i;climbStaris_CreateTree(step, left);}}}/** * Description: 打印楼梯的走法,即遍历整个树 *  * @param root *            需要遍历的父节点 * @param rootPath *            需要遍历的父节点相对于树根节点的路径 */public static void printPath_ergodicTree(Node root, String rootPath) {// 若传入的父节点即为叶子节点,则直接打印其路径,完毕后回,遍历下一个节点if (root.children.size() == 0) {total++;System.out.println(rootPath);return;}// 若传入的父节点有子节点,则递归遍历其子节点for (int i = 0; i < root.children.size(); i++) {Node child = root.children.get(i);String childPath = rootPath + "-" + String.valueOf(child.val);// 递归处理其子节点printPath_ergodicTree(child, childPath);}}public static void main(String[] args) {Node root = new Node();climbStaris_CreateTree(root, 20);printPath_ergodicTree(root, "start");System.out.println(total);}}/** * Description: 定义树的节点 *  * @author karl */class Node {int val;List<Node> children = new ArrayList<Node>();Node() {}Node(int val) {this.val = val;}}

斐波那契数列实现：

f(n) = 1 (n = 1);    = 2 (n = 2);  = 4 (n = 3);  = f(n-3) + f(n-2) + f(n-1) (n>3);

 

package com.csdn.question;import java.util.ArrayList;import java.util.List;public class Fibonacci {private static int total = 0;public static void main(String[] args) {goUp(3);System.out.println(total);}public static void goUp(int n) {List<String> fibList = new ArrayList<String>();fibonacci(n, fibList);}private static void fibonacci(int n, List<String> fibList) {if (n == 0) {for (String str : fibList) {System.out.print(str + " ");}total++;System.out.println();} else if (n < 0) {return;} else {int num = fibList.size();fibList.add("1");fibonacci(n - 1, fibList);fibList = fibList.subList(0, num);fibList.add("2");fibonacci(n - 2, fibList);fibList = fibList.subList(0, num);fibList.add("3");fibonacci(n - 3, fibList);fibList = fibList.subList(0, num);}}}