70. Climbing Stairs

题意：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：这题其实用排列组合的方法也可以解，但这提的本质其实就是菲波那切数列，r[i+2]其实是由r[i+1]和r[i]分别走一步和两部构成的，所以r[i+2] = r[i+1]+r[i]，这样的话解法就很多了。

C++:这是我最开始的解法，但是时间超了。。。：

class Solution {public:    void recur(int cur, int n, int &result){        if (cur > n)            return;        else if (cur == n){            result += 1;            return;        }        else{            recur(cur + 1, n, result);            recur(cur + 2, n, result);        }    }    int climbStairs(int n) {        int result = 0;        recur(0, n, result);        return result;    }};

Python：当然，可以不用设置数组，直接用常数空间求解：

def climbStairs(self, n):    a = b = 1    for _ in range(n):        a, b = b, a + b    return a