[LeetCode]Climbing Stairs

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?

n楼的走法是第n-1楼走一步或者n-2楼走两步，所以有公式f(n)=f(n-1)+f(n-2)。就是斐波那契数列。

这种有规律的题有个共同解法，建立一个数组或vector记录每一个解，很好理解，不过空间复杂度为O(n)。

class Solution {
public:
    int climbStairs(int n) {
        vector<int> f(n+1,0);
        f[1]=1;
        f[2]=2;
        for(int i=3;i<=n;i++)
        {
            f[i]=f[i-2]+f[i-1];
        }
        return f[n];
    }
};

使用迭代可以将空间复杂度降为O(1)。

class Solution {
public:
    int climbStairs(int n) {
        vector<int> f(n+1,0);
        int pre = 0;
        int cur =1;
        for(int i=1;i<=n;i++)
        {
            int temp = cur;
            cur = cur+pre;
            pre = temp;
        }
        return cur;
    }
};