在深信服组期间学习了C++的使用方法，就打算通过刷Leetcode来提高自己的算法能力以及对C++的掌握能力

递归和动规的典型问题，也是迅速ac了

Leetcode传送门

Climbing Stairs [Difficulty: Easy]

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:
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    Input: 2
    Output: 2
    Explanation: There are two ways to climb to the top.
    1. 1 step + 1 step
    2. 2 steps

    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
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class Solution {
public:
    int climbStairs(int n) {
        
    }
};

我的思路

  • 第一想法是递归，写出第一版代码如下

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    class Solution {
    public:
        int climbStairs(int n) {
            if (n == 1) return 1;
            if (n == 2) return 2;
            return climbStairs(n-1) + climbStairs(n-2);
        }
    };

  • 显然这代码超时了，因为递归过程中很多步骤会被重复计算2次，所以最好是把中间值存起来，这时候动规的优势就体现出来了

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class Solution {
public:
    int climbStairs(int n) {
        if (n == 1) return 1;
        if (n == 2) return 2;
        int results[n+1];
        results[1] = 1;
        results[2] = 2;
        for (int i=3; i<=n; ++i)
            results[i] = results[i-1] + results[i-2];
        return results[n] ;
    }
};