關於青蛙跳的問題解答

提問：一隻青蛙一次可以跳上 1 級臺階，也可以跳上 2 級。求該青蛙跳上一個 n 級的臺階總共有多少種跳法。

由於打印了青蛙跳的步驟，所以運行速度非常慢。

import java.util.Date;

public class Frag {

	public static int STEP;
	public static int TOT = 0;
	public static String tmpprint;

	public static void goNextJump(int currentStep, int length, String print) {
		if (currentStep == STEP) {
			if (!print.equals(tmpprint)) {
				TOT = TOT + 1;
				 System.out.println(print + "(TOT=" + TOT + ")");
				tmpprint = print;
			}
			return;
		} else {
			if (currentStep + length <= STEP) {
				print = print + "-->" + length;
				currentStep = currentStep + length;
				goNextJump(currentStep, 1, print);
				goNextJump(currentStep, 2, print);
			}
			return;
		}
	}

	public static int getTot(int step) {
		STEP = step;
		if (step == 0) {
			return 0;
		}
		if (step == 1) {
			return 1;
		}
		String print = "0";
		goNextJump(0, 1, print);
		goNextJump(0, 2, print);
		return TOT;
	}

	public static void main(String[] args) {
		Date start = new Date();
		System.out.println(getTot(20));
		Date end = new Date();
		System.out.println(end.getTime() - start.getTime());
	}

}

  我的同事Vivi給出了強大的答案

/**
 * @description 一隻青蛙一次可以跳上 1 級臺階，也可以跳上2 級。求該青蛙跳上一個n 級的臺階總共有多少種跳法
 * @author vivi_zhao
 * @date Dec 30, 2011
 */
public class Frog {

	public long[] a;
	public boolean[] isExist;
	
	void init(int n) {
		a = new long[n + 1];
		isExist = new boolean[n + 1];
	}

	long hoopWithMemory(int n) {
		if (n < 0)
			return 0;
		if (n == 0)
			return 1;
		if(!isExist[n]) {
			isExist[n] = true;
			return a[n] = hoopWithMemory(n - 2) + hoopWithMemory(n - 1);
		}
		return a[n];
	}

	long hoopWithoutMemory(long n) {
		if(n < 0)
			return 0;
		if(n == 0)
			return 1;
		return hoopWithoutMemory(n - 1) + hoopWithoutMemory(n - 2);
	}
	
	public static void main(String[] args) {
		Frog frog = new Frog();
		int count = 50;
		long start = System.currentTimeMillis();
		//hoop with memory
		frog.init(count);
		start = System.currentTimeMillis();
		System.out.println(frog.hoopWithMemory(count));
		long end = System.currentTimeMillis();
		System.out.println(frog.hoopWithoutMemory(count));
		long end2 = System.currentTimeMillis();
		System.out.println("(hoop with memory = " + (end - start) + ") << (hoop without memory = " + (end2 - end) + ")");
	}
}