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?

簡單題目，相當於fibonacci數列問題，難點就是要會思維轉換，轉換成為遞迴求解問題，多訓練就可以了。

所以這種型別的題目相對於沒有形成遞迴邏輯思維的人來說，應該算是難題。

我的想法是：

每次有兩種選擇，兩種選擇之後又是各有兩種選擇，如此迴圈，正好是遞迴求解的問題。

寫成遞迴程式其實非常簡單，三個語句就可以：

int climbStairsRecur(int n) {
		if (n == 1) return 1;
		if (n == 2) return 2;
		return climbStairsRecur(n-1) + climbStairsRecur(n-2);
	}

int climbStairs(int n)
	{
		vector<int> res(n+1);
		res[0] = 1;
		res[1] = 1;
		for (int i = 2; i <= n; i++)
		{
			res[i] = res[i-1] + res[i-2];
		}
		return res[n];
	}
 

int climbStairs2(int n)
	{
		vector<int> res(3);
		res[0] = 1;
		res[1] = 1;
		for (int i = 2; i <= n; i++)
		{
			res[i%3] = res[(i-1)%3] + res[(i-2)%3];
		}
		return res[n%3];
	}

//2014-2-10 update
	int climbStairs(int n)
	{
		if (n < 4) return n;
		int a = 2, b = 3, c = 5;
		for (int i = 5; i <= n; i++)
		{
			a = c;
			c = b+c;
			b = a;
		}
		return c;
	}