Dynamic Programming Topic (II.)--Jumping steps

Dynamic Programming Topic (II.)--Jumping steps

1. Title Description

A step has a total of n levels, if you can jump 1 levels at a time, you can jump 2 levels.

How many total hops are in total, and the time complexity of the algorithm is analyzed.

2. Recursive method

For this dynamic planning problem, we have two steps to think of:

• If we jump 1 levels, then the rest of the jump is f (n-1);
• If we jump 2 levels, then the rest of the jump is f (n-2);

this time we will be able to implement the recursive, slow! We also need to jump out of the recursive conditions, this is also essential!
For This topic, the condition of jumping out of recursion is when n==1, 2, we return 1, 2.
Why is it? Because when N=1, there is only one kind of jumping method, when n=2, there are two kinds of jumping method. Then there are the initial values, and there are only two cases (either jumping 1 levels or jumping 2 levels), so This is a Fibonacci sequence problem!


The code is as follows:

#include <iostream>using namespace std; int findsolution (int N) {int result[3]={0, 1, 2};   Initial condition if (n<=2) return result[n]; Return Findsolution (N-1) +findsolution (N-2);   Recursive implementation}int Main () {int n=20; int num=findsolution (N); Cout<<num<<endl;return 0;}

3. Non-recursive mode

The non-recursive way is also must be mastered!

If this is a Fibonacci sequence problem, then the non-recursive way to achieve it is relatively easy ~

As follows:

#include <iostream>using namespace std; int findsolution (int N) {int fab[3]={1,1}, if (n<2) return 1; for (int i=2; i<=n; i++) {fab[2]=fab[0]+fab[1];   It is necessary to pay attention to the calculation rule fab[0]=fab[1]; FAB[1]=FAB[2]; }return fab[2]; }int Main () {int n=20; int num=findsolution (N); Cout<<num<<endl;return 0;}

It should be pointed out that the problem of the non-recursive method of the calculation of the rule is better, then the "Dynamic Planning (A)" in the article is not very good to think, we must pay attention to the different problems of the calculation rules!

In addition, if you can jump 3 levels, then the implementation of the idea is the same, just need to replace the initial number of values can be ~

Dynamic Programming Topic (II.)--Jumping steps