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
这道题粗看很难解,分类能分到吐血,而且不好循环,但是写几个就能发现规律了。
1->1 2->2 3->3 4->5 5->8
典型的斐波拉契数列。
我一开始代码为
class Solution {
public int climbStairs(int n) {
if(n==1)
return 1;
if(n==2)
return 2;
else
return climbStairs(n-1)+climbStairs(n-2);
}
}
但是超时了,因为这种递归会重复计算。
后来用了数组来存储。
class Solution {
public int climbStairs(int n) {
if(n==1)
return 1;
int[]res=new int[n+1];
res[1]=1;
res[2]=2;
for(int i=3;i<n+1;i++)
{
res[i]=res[i-1]+res[i-2];
}
return res[n];
}
}
附斐波拉契数列一种很好的求法。
int Fib2(int n,int a,int b)//a表示第一个数的值,b表示第二个数的值
{
if(n<=2)
return b;
return Fib2(n-1,b,a+b);
}