Java高频经典面试题（第一季）五：递归与迭代

package will01;

import org.junit.Test;

public class TestStep {
    @Test
    public void test(){
        long start = System.currentTimeMillis();
        System.out.println(f(30));
        long end = System.currentTimeMillis();
        System.out.println("time : "+ (end - start));
        
    }
    
    
    
    //实现f(n):求 n步 台阶，一共有 几种 走法
    public int f(int n ){
        if(n < 1){
            throw new IllegalArgumentException(n + "不能小于1 ");
        }
        if(n == 1 || n == 2){
            return n ;
        }
        return f(n - 2)+ f( n - 1);
    }

}

package will01;

import org.junit.Test;

public class TestStep2 {
    
    @Test
    public void test(){
        long start = System.currentTimeMillis();
        System.out.println(loop(40));
        long end = System.currentTimeMillis();
        System.out.println("time : "+ (end - start)); // < 0ms
        
    }
    
    
    public int loop(int n){
        if(n < 1){
            throw new IllegalArgumentException(n + "不能小于1 ");
        }
        if(n == 1 || n == 2){
            return n;
        }
        int one = 2; // 初始化为走到第二台阶的走法
        int two = 1; // 初始化为走到第一台阶的走法
        int sum = 0;
        
        for(int i = 3; i <= n ; i++){
            //最后跨两步 + 最后跨一步 的走法
            sum = two + one ;
            two = one;
            one = sum;
        }
        
        return sum;
        
    }

}