【LeetCode】Climbing Stairs (2 solutions)

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?

这是最简单的DP了，递推公式为f(n)=f(n-1)+f(n-2)

因此建立vector存放中间结果即可。

class Solution
{
public:
    int climbStairs(int n)
    {
        vector<int> v;

        for(int i = 0; i <= n; i ++)
        {
            if(i == 0)
                v.push_back(1);
            else if(i == 1)
                v.push_back(1);
            else
                v.push_back(v[i-2]+v[i-1]);
        }

        return v[n];
    }
};

其实可以改进一下减少空间复杂度。

每个元素仅依赖于前两个元素，因此没有必要使用vector保存所有中间结果，只需要保存前两个。

class Solution
{
public:
    int climbStairs(int n)
    {
        if(n <= 1)
            return 1;
        int pre1 = 1;    //n==0
        int pre2 = 1;    //n==1
        for(int i = 2; i <= n; i ++)
        {
            int cur = pre1+pre2;
            pre1 = pre2;
            pre2 = cur;
        }

        return pre2;
    }
};