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
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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12Input: 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
1 | class Solution { |
我的思路
第一想法是递归,写出第一版代码如下
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8class 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
1 | class Solution { |