标签:io for 问题 ar res return public 表达式
问题描述:
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?
解题思路:
这是一个典型的斐波那契数列,对于有n步的台阶:
1、如果先跳1步,还剩余n-1步,那么这就变成n-1步台阶跳法数目;
2、如果先跳2步,还剩余n-2步,这就变成n-2步台阶跳法数目。
所以经典的表达式:f(n) = f(n-1) + f(n-2)
class Solution {
public:
int climbStairs(int n) {
/*f(n)=f(n-1)+f(n-2)*/
if(n==0||n==1) return 1;
int stepOne=1,stepTwo=1;
int result=0;
for(int i=2;i<=n;i++){
result=stepOne+stepTwo;
stepTwo=stepOne;
stepOne=result;
}
return result;
}
};Climbing Stairs,布布扣,bubuko.com
标签:io for 问题 ar res return public 表达式
原文地址:http://blog.csdn.net/wan_hust/article/details/38307563