动态规划--爬楼梯

1、题目:

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

2、解答: 假设第n步,你可以由n-1步,走1步到达第n步,n-2,走2步到达第n步。

3、代码:

C++代码

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

python代码

class Solution:
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n < 1:
            return 0
        
        f = [1 for i in range(n+1)]
        
        for i in range(2,len(f)):
            f[i] = f[i-1] + f[i-2]
        
        return f[n]

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转载自blog.csdn.net/qq_31307013/article/details/80575944