Climbing Stairs_LeetCode

Description:

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

 

 解题思路：

可以这样理解：比如当下台阶数为n，则到达当前台阶有两种可能性：

1.从n-1号台阶走1个台阶到达；

2.从n-2号台阶走2个台阶到达。

起初，用递归做，但是超时了，因此需要优化算法，根据上面的思路可以利用简单的动态规划完成。

列出动态规划的式子：dp[n] = dp[n-1] + dp[n-2];

以及一些基本的情况：

当 n <= 0时，result = 0;

当 n == 1时，result = 1;

当 n == 2时，result = 2;

 

代码：

class Solution {
public:
    int climbStairs(int n) {
        vector<int> dp(n+1);
        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];
    }
};