leetcode 70. Climbing Stairs QuestionEditorial

/*
70. Climbing Stairs  QuestionEditorial Solution  My Submissions
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?

解题思路：
由于走到第n阶的最后一步，可以走一步或者两步，那么第n阶的步数等于第n-1阶步数+n-2阶步数。
dp[n] = dp[n-1]+dp[n-2]
*/

#include <iostream>
#include <string>
#include <vector>

using namespace std;

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

        return dp[n];
    }

    int climbStairs1(int n)
    {
        if (n < 3)
            return n;
        int pre1 = 2;   //前一个的数字(n-1)的step
        int pre2 = 1;   //n-2的step
        int ret = 0;
        int tmp = 0;
        for (int i = 3; i <= n; ++i)
        {
            ret = pre1 + pre2;
            //cout << ret << endl;
            tmp = pre1; //更新pre1和pre2
            pre1 = ret;
            pre2 = tmp;
        }
        //cout << "---------------" << endl;
        return ret;
    }
};

void test()
{
    Solution sol;
    cout << sol.climbStairs1(10) << endl;
}

int main()
{
    test();

    return 0;
}