题目描述
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
动态规划,等价于斐波那契数列。
具体实现
class Solution {
public:
int climbStairs(int n) {
int a = 1, b = 1;
for (int i = 1; i < n; i++) {
b = a + b;
a = b - a;
}
return b;
}
};
本文探讨了一个经典的爬楼梯问题,即用动态规划的方法求解到达楼顶的不同方式数量。文章给出了具体的例子,并通过斐波那契数列的性质提供了一种简洁的解决方案。
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