[Sword refers to Offer] 2. abnormal jump steps, sword refers to offer

[Sword refers to Offer] 2. abnormal jump steps, sword refers to offer

Question

A frog can jump to level 1 or Level 2 at a time ...... It can also jump to n levels. Find the total number of hops that the frog jumps to an n-level step.

Ideas

Use Fib (n) to represent the number of hops on the n-step. Set Fib (0) to 1;

When n = 1, there is only one hop method, that is, level 1 hop, that is, Fib (1) = 1;

When n = 2, there are two hop modes: first hop and second hop, that is, Fib (2) = Fib (1) + Fib (0) = 2;

When n = 3, there are three ways to skip. After jumping out of the first step for the first time, there is also the Fib (3-1) Middle hop method. After jumping out of the second step for the first time, there is also the Fib (3-2) Middle hop method in the back. After the first jump out of the third step, there is also the Fib (3-3) Middle hop method in the back, that is, Fib (3) = Fib (2) + Fib (1) + Fib (0) = 4;

When n = n, there are n skip methods. After jumping out of the first step for the first time, there is also the Fib (n-1) Middle hop method. After jumping out of the second step for the first time, there is a Fib (n-2) in the back of the jump .......................... After the first jump out of the n step, there is a Fib (n-n) in the jump method, that is, Fib (n) = Fib (n-1) + Fib (n-2) + Fib (n-3) + .......... + Fib (n-n) = Fib (0) + Fib (1) + Fib (2) + ....... + Fib (n-1) and because Fib (n-1) = Fib (0) + Fib (1) + Fib (2) + ....... + Fib (n-2) So Fib (n) = 2 * Fib (n-1) n> = 2

To sum up:

Code

/* ------------------------------------- * Date: 2015-07-19 * Author: SJF0115 * Title: 2. abnormal jump steps * Web site: http://www.nowcoder.com/books/coding-interviews/22243d016f6b47f2a6928b4313c85387? Rp = 1 * result: AC * Source: offoff * blog: ----------------------------------------- */# include <iostream> using namespace std; class Solution {public: int jumpFloorII (int number) {if (number <= 0) {return 0;} // if else if (number = 1) {return 1 ;} // else return 2 * jumpFloorII (number-1) ;}}; int main () {Solution s; int number = 5; cout <s. jumpFloorII (number) <endl; return 0 ;}

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