问题描述：
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?
提示：Dinamic Programming，动态规划，这是个斐波那契数列问题。
方法一：
//找规律，设n个台阶的方法数为f(n)
//f(1) = 1;f(2) = 2; .... ,从第n-1走到第n台阶，有两种方式：一步到或者两步到。
//因此，f(n) = f(n-1) + f(n-2); (n >= 3)
//递归运算，时间复杂度太高
public static int climbs(int n){
if(n == 0) return 0; if(n == 1)
return 1;
if(n == 2)
return 2;
return climbs(n - 1) + climbs(n - 2); //分类
}
方法二：时间复杂度O(n)，空间复杂度O(1)
public int climbStairs(int n) {
if(n == 0)
return 0;
int prev = 0; //prev = f(n-2)
int cur = 1; //cur = f(n-1)
for(int i = 1; i <= n ; ++i){
int tmp = cur;
cur = prev + cur; // f(n) = f(n-2) + f(n-1)
prev = tmp;
}
return cur;
}