爬楼梯问题,描述如下:
You are climbing a staircase. 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.
题目大概意思就是爬楼梯每次能爬一层或两层,求 爬到第n层共有多少种爬法。这是一道动态规划问题,设爬到第i层共有d[i]种爬法,因为一次能爬一层或两层,所以爬到第i层可以从i-1层爬一层和i-2层爬两层,即d[i] = d[i-1]+d[i-2],且d[1]=1,d[2]=2,这点和斐波那契数列完全一致,代码如下。
class Solution {
public:
int climbStairs(int n) {
int s1 = 1;
int s2 = 1;
for(int i = 1;i < n;++i)
{
s2 = s1+s2;
s1 = s2-s1;
}
return s2;
}
};