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?
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刚开始用递归做的,递归比较麻烦的在于没有用空间存储吧,所以时间比较慢。
其实思路不能再简单了,就是斐波那契数列,改了一下而已就是F(1)=1,F(2)=2,F(n)=F(n-1)+F(n-2);
和那天当当网的那个也很像,发现这种题目其实画二叉树的话比较容易找到思路。
class Solution { public: int climbStairs(int n) { int array[n]; array[0]=1; array[1]=2; for(int i=2;i<n;i++){ array[i]=array[i-1]+array[i-2]; } return array[n-1]; } };