Climbing Stairs
Climbing Stairs
https://leetcode.com/problems/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?
算法描述
(1)这道题其实是一道fibonacci数列题。
当n=1时,f(n)=1;当n=2时,f(n)=2(有1+1,2这两种方式);当n>=3时,有f(n)=f(n-1)+f(n-2)
(2)但是实现时如果直接使用递归,就会超时,所以这里使用循环
(3)用f0来记录f(n-2),用f1来记录f(n-1),按照之前的公式f(n)=f(n-1)+f(n-2),即得f2=f0+f1;然后更新f0和f1,使得这两个记录都往前移动一位,即f0=f1,f1=f2;
一共进行n-1次,返回f2
程序代码
public class Solution {
public int climbStairs(int n) {
int f0 = 1;
int f1 = 1;
int f2 = n;
n--;
while (n-- > 0) {
f2 = f0 + f1;
f0 = f1;
f1 = f2;
}
return f2;
}
}
算法2-尾递归
对于这道题,记得之前有本书提到过尾递归的思想,所以这里应用了一把尾递归,不过单就这道题看来,其实就跟循环的中心思想是一样的,都是记录做过的两个状态。
程序代码2
public class Solution {
public int climbStairsTail(int n, int acc, int acc2) {
if (n == 0) {
return acc;
}
return climbStairsTail(n - 1, acc2, acc + acc2);
}
public int climbStairs(int n) {
if (n == 0 || n == 1) {
return n;
}
return climbStairsTail(n, 1, 1);
}
}
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