爬楼梯问题

简述：

需要爬N层的楼梯，有max种爬楼梯方式，需要求算总共爬到N层 有几种方法

1. 回溯是比较直观的，就是一层一层向上叠加，走到顶层就是一种方法，得到结果

2. 一维动态规划的话， 就是用一个数组保留走过的楼层，比如最大步伐是3， 然后我现在考虑走到第10层有几种走法

就是 resultArray[10] = resultArray[ 10 - 1] + resultArray[10 - 2] + resultArray[10 - 3]  三者之和

3. 3解法中优化了2中的空间复杂度，用取余的方式实现f(n) = f(n - 1) + f(n - 2) + ... + f(n - k)

例子：

爬4层楼梯 ， 则N=4， 有3种步伐，分别为1步，2步，3步

则总共方法7种，一次为

1 + 1 + 1 + 1

1 + 1 + 2

1 + 2 + 1

1 + 3

2 + 1 + 1

2 + 2

3 + 1

算法实现1(回溯法）：

使用函数压栈， 不断累加当前楼梯的高度，深度探测是否已经爬到顶了，如果爬到顶了，那么count++，没有那么继续爬，并把步子压到stack中记录步伐顺序

package dynamic_programming;import java.io.BufferedReader;import java.io.IOException;import java.io.InputStream;import java.io.InputStreamReader;import java.util.ArrayList;import java.util.Stack;public class ClimbTheFloor {private static int count = 0;private static Stack<Integer> stack = new Stack<Integer>();private static ArrayList<String> resultList = new ArrayList<String>();public static void main(String[] args) {InputStream inputStream = System.in;  InputStreamReader inputStreamReader = new InputStreamReader(inputStream);     BufferedReader bufferReader = new BufferedReader(inputStreamReader);  //Get limited heightSystem.out.print("Please Input the Max height: ");Integer max = null;try{    max = Integer.parseInt(bufferReader.readLine());  }catch(IOException e){    e.printStackTrace();    }  //Get the max stepSystem.out.print("Please Input the limited step: ");Integer limit = null;    try{    limit = Integer.parseInt(bufferReader.readLine());  }catch(IOException e){e.printStackTrace();    }Climb(0, max, limit);System.out.println("There are " + count + " methods !");for(String record : resultList)System.out.println(record);}public static void Climb(int height, int max, int limit){if(height == max){count++;resultList.add(stack.toString());}else{for(int i = 1;i <= limit; i++)if(height + i <= max){stack.push(i);Climb(height + i, max, limit);stack.pop();}}}}

输出：

2） 动态规划实现：

package dynamic_programming;import java.io.BufferedReader;import java.io.IOException;import java.io.InputStream;import java.io.InputStreamReader;public class ClimbTheFloor2 {public static void main(String[] args) {InputStream inputStream = System.in;  InputStreamReader inputStreamReader = new InputStreamReader(inputStream);     BufferedReader bufferReader = new BufferedReader(inputStreamReader);  //Get limited heightSystem.out.print("Please Input the Max height: ");Integer max = null;try{    max = Integer.parseInt(bufferReader.readLine());  }catch(IOException e){    e.printStackTrace();    }  //Get the max stepSystem.out.print("Please Input the limited step: ");Integer limit = null;    try{    limit = Integer.parseInt(bufferReader.readLine());  }catch(IOException e){e.printStackTrace();    }System.out.println("Number of Methods: " + CountMethods(max, limit));}/* * @ Height : the height of the destination     * @ max : the max number of steps you can climb upstairs once */private static int CountMethods(int max, int limit){int resultArray[] = new int[max + 1];resultArray[0] = 1;for(int i = 1;i <= max;i++){int sum = 0;for(int j = 1;j <= limit;j++){if(i - j < 0)break;else{sum += resultArray[i - j];}}resultArray[i] = sum;}return resultArray[max];}}

输出：

3） 优化了2）中的空间复杂度， 将空间降到了O(k) , 其中k为可以走的步数的种类

package dynamic_programming;import java.io.BufferedReader;import java.io.IOException;import java.io.InputStream;import java.io.InputStreamReader;public class ClimbTheFloor3 {public static void main(String[] args) {InputStream inputStream = System.in;InputStreamReader inputStreamReader = new InputStreamReader(inputStream);BufferedReader bufferedReader = new BufferedReader(inputStreamReader);Integer limit = null;try{System.out.print("Please Input Limited step:");limit = Integer.parseInt(bufferedReader.readLine());}catch(IOException e){e.printStackTrace();}Integer max = null;try{System.out.print("\nPlease Input Max Height: ");max = Integer.parseInt(bufferedReader.readLine());}catch(IOException e){e.printStackTrace();}ClimbTheFloor3 obj = new ClimbTheFloor3();System.out.print("\nMethods: " + obj.CountMethods(max, limit));}private int CountMethods(int max, int limit){if(max < 0 || limit < 0)throw new IllegalArgumentException("Wrong arguments! ");int step[] = new int[limit];for(int i = 0; i < limit; i++)step[i] = 1;for(int i = 1; i <= max; i++){int sum = 0;for(int j = 0; j < limit; j++){if(i - (j + 1) == 0){sum++;break;}sum += step[(i - (j + 1)) % limit];}step[(i - 1) % limit] = sum;}return step[(max - 1) % limit];}}

输出：