[Leetcode] Climbing Stairs

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?

题意：台阶总数为n，每次爬1阶或者2阶层，求爬完台阶的不同的方法数。

第一次爬1阶，则剩下n-1阶；

第一次爬2阶，则剩下n-2阶；

递推表达式为：

F(n) = F(n-1) + F(n-2)，F(1) = 1, F(2) = 2;

由于F(n) = F(n-1) + F(n-2) = (F(n-2) + F(n-3)) + (F(n-3) + F(n-4))=.....

由于计算结果没有保存，越往后，重复计算越多，考虑使用动态规划，将F(n-1)和F(n-2) 保存下来。

public class Solution {
public int climbStairs(int n) {

if (n == 1 || n == 2) {
return n;
}

int pre = 2, beforePre = 1;
int curr = 3;
for (int i = 3; i <= n; ++i) {
curr = pre + beforePre;
beforePre = pre;
pre = curr;
}
return curr;
}
}