Climbing Stairs

Number of 1 Bits

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?

思路

climb(n) = climb(n-1)+climb(n-2)

But

如果你写出这样的代码

class Solution {
public:
int climbStairs(int n) {
if(n==1)
return 1;
else if(n==2)
return 2;
else
return climbStairs(n-1)+climbStairs(n-2);
}
};

结果是，超时！！！！

这是斐波拉契数列，n-1可以用到n-2的值，不需要重复计算

代码

class Solution {
public:
int climbStairs(int n) {
if(n==0||n==1||n==2)
return n;
/* 用数组
int res
;
res[1]=1;
res[2]=2;
for(int i= 3;i<=n;i++)
res[i]= res[i-2]+res[i-1];
return res
;
*/

int a =1,b=2,res;

for(int i = 3;i<=n;i++)
{
res = a+b;
a=b;
b=res;
}
return res;
}
};