动态规划系列 Leetcode 70. 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?

Note: Given n will be a positive integer.

 

Example 1:

Input: 2
Output:  2
Explanation:  There are two ways to climb to the top.

1. 1 step + 1 step
2. 2 steps

 

Example 2:

Input: 3
Output:  3
Explanation:  There are three ways to climb to the top.

1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

这个一道基本动态规划题目，做动态规划题目有四个步骤：
1）确定原问题和子问题
2）确定状态
3）确认边界状态的值
4）确定状态转移方程

#include <stdio.h>

#include <vector>
class Solution {
public:
    int climbStairs(int n) {
        std::vector<int> dp(n + 3, 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 main(){
    Solution solve;
    printf("%d\n", solve.climbStairs(3));    
    return 0;
}