[10.2] skip steps prove safety

Title Description

A frog can jump on a Class 1 level, you can also hop on level 2. The frog jumped seeking a total of n grade level how many jumps.

analysis

With Fibonacci number, 1. 2. recursive solving dynamic programming

public class JumpFloor {

    public static void main(String[] args) {
        System.out.println(jump1(3));
        System.out.println(jump1(5));

        System.out.println(jump2(3));
        System.out.println(jump2(5));
    }

    //1.递归求解
    public static int jump1(int n) {
        if (n <= 2) return n;
        return jump1(n - 1) + jump1(n - 2);
    }

    //2.动态规划
    public static int jump2(int n) {
        if (n <= 2) return n;
        int result = 0;
        int pre2 = 1;
        int pre1 = 2;
        for (int i = 3; i <= n; i++) {
            result = pre1 + pre2;
            pre2 = pre1;
            pre1 = result;
        }
        return result;
    }

}

Reproduced in: https: //www.jianshu.com/p/5a808473e855