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:
利用DP的方法,一個臺階的方法次數爲1次,兩個臺階的方法次數爲2個。n個臺階的方法可以理解成上n-2個臺階,然後2步直接上最後一步;或者上n-1個臺階,再單獨上一步。
公式是S[n] = S[n-1] + S[n-2] S[1] = 1 S[2] = 2
該題最好不使用遞歸,因爲遞歸耗費的時間太長,所以不能使用遞歸。
*/
#include "stdafx.h"
int climbStairs(int n);
int _tmain(int argc, _TCHAR* argv[])
{
int n = climbStairs(44);
return 0;
}
int climbStairs(int n)
{
if (n == 0)
return 0;
if (n == 1)
return 1;
if (n == 2)
return 2;
int a = 1, b = 2;
int sum = 0;
if (n > 2)
{
for (int i = 3; i <= n; i++)
{
sum = a + b;
a = b;
b = sum;
}
}
return sum;
}