版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/bless2015/article/details/88126948
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?
Note: Given n will be a positive integer.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
- 1 step + 1 step
- 2 steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.- 1 step + 1 step + 1 step
- 1 step + 2 steps
- 2 steps + 1 step
class Solution {
public int climbStairs(int n) {
if(n<2){
return n;
}
int f[] = new int[n+1];
f[0]=1;
f[1]=1;
for (int i = 2; i <= n; i++) {
f[i]= f[i-1]+f[i-2];
}
return f[n];
}
}
这道题其实是求斐波那契数列,这种解法存在优化,因为f[i-1]和f[i-2]在计算过程中确实是被重复计算了。优化的方法又叫备忘录法,牺牲一定的空间来保存每个中间解。