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leetcode 70 Climbing Stairs

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流川疯
发布2019-01-18 15:12:38
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发布2019-01-18 15:12:38
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文章被收录于专栏:流川疯编写程序的艺术

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?

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Dynamic Programming

这个题面试题还是比较常见的

讨论的帖子:

https://leetcode.com/discuss/2809/easy-solutions-for-suggestions

原理:

This problem is a Fibonacci problem. F(n)=F(n-1)+F(n-2); Solving this problem by recursion ,we will do a lot of same recursion. Example: F(10)=F(9)+F(8); F(9)=F(8)+F(7); we calculate F(8) twice,when n is large,this will increase as a rate of n's exponent.

So a more efficient way to solve this problem is from Bottom to Top. Calculate F(0) ,F(1); then F(2).........

人生ac最快的代码:

代码语言:javascript
复制
class Solution {
public:
    int climbStairs(int n)
    {
        
        int stepone = 0;
        int steptwo = 1;
        int sum = 0;
        
        for(int i = 0;i<n;i++)
        {
            sum = stepone + steptwo;
            stepone = steptwo;
            steptwo = sum;
            
            
        }
        
        return sum;
    }
};

DP算法求解:

https://leetcode.com/discuss/16275/my-dp-solution-using-c-4-ms

简洁的代码:

https://leetcode.com/discuss/31848/1ms-in-c-and-2ms-in-c-optimal-space-no-temp-value

递归:

https://leetcode.com/discuss/28383/simple-and-clear-2ms-solution-in-c-without-recursion

python:

https://leetcode.com/discuss/25378/this-is-essentially-a-fibonacci-sequence

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