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?
设f(n)表示爬n阶楼梯的不同方法数,为了爬到第n阶楼梯,有两个选择:
- 从第
n-1阶前进1步; - 从第
n-1阶前进2步;
因此,有f(n)=f(n-1)+f(n-2)。
这是一个斐波那契数列。
方法1,递归,太慢;方法2,迭代。
方法3,数学公式。斐波那契数列的通项公式为
// Climbing Stairs
// 迭代,时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
int climbStairs(int n) {
int prev = 0;
int cur = 1;
for(int i = 1; i <= n ; ++i){
int tmp = cur;
cur += prev;
prev = tmp;
}
return cur;
}
};{% codesnippet "./code/climbing-stairs-2."+book.suffix, language=book.suffix %}{% endcodesnippet %}