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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.

思路：假设距离走到第n阶台阶只差一步，则可能是走到了n-1阶或者n-2阶。n-1阶到第n阶dp[n-2]，以及n-2阶到第n阶有dp[n-3]种情况。且n-1阶到第n阶，n-2阶到第n阶，完全独立。dp[n-1]=dp[n-2]+dp[n-3]。

class Solution {public:    int climbStairs(int n) {        vector<int> dp(n);//给容器开辟空间dp[0]=1;dp[1]=2;for(int i=2;i<n;i++){dp[i]=dp[i-1]+dp[i-2];}return dp[n-1];    }};