/**
*
* @author gentleKay
* You are climbing a stair case. It takes n steps to reach to the top.
* Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
*
* 你在爬楼梯。它需要N个步骤才能到达顶部。
* 每次你可以爬1到2步。你能用多少种不同的方式爬到顶峰？
*/
这是一题动态规划的题目，可以和这道题一起来看：
https://www.cnblogs.com/strive-19970713/p/11262614.html
方法一：（动态规划 空间和时间复杂度都为O(n)）
/**
*
* @author gentleKay
* You are climbing a stair case. It takes n steps to reach to the top.
* Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
*
* 你在爬楼梯。它需要N个步骤才能到达顶部。
* 每次你可以爬1到2步。你能用多少种不同的方式爬到顶峰？
*/
public class Main22 {
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(Main22.climbStairs(1));
}
public static int climbStairs(int n) {
int[] a = new int[n+1];
a[0] = 1;
a[1] = 1;
for (int i=2;i<a.length;i++) {
a[i] = a[i-1] + a[i-2];
}
return a[n];
}
}
方法二：（递归 在letcode会超时错误）
/**
*
* @author gentleKay
* You are climbing a stair case. It takes n steps to reach to the top.
* Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
*
* 你在爬楼梯。它需要N个步骤才能到达顶部。
* 每次你可以爬1到2步。你能用多少种不同的方式爬到顶峰？
*/
public class Main22 {
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(Main22.climbStairs(1));
}
public static int climbStairs(int n) {
if (n == 1) {
return 1;
}
if (n == 2) {
return 2;
}
return climbStairs(n-1) + climbStairs(n-2);
}
}