Leet Code OJ 70. Climbing Stairs [Difficulty: Easy]

题目：
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?

翻译：
你在爬一个楼梯。到达顶部有n级阶梯。
每次你可以选择爬一级或者二级。在多少不同的方式去到达顶部？

分析：
当n=1，无疑只有一种方式，s=1。
n=2，s=2。
n=3，s=3。
n=4，s=5。
我们发现这个实际是一个斐波那契数列。第一反应是递归实现，f(n)=f(n-1)+f(n-2)，但是递归实现，有多次重复计算，比如在计算f(n-1)时，需要使用f(n-2)+f(n-3)，但是这个时候另一个递归调用已经去算了f(n-2)，相当于f(n-2)被计算了2次，这种重复计算的情况普遍存在，将会浪费大量计算时间。很自然的想到反过来操作，递归是从目标算到基础值，而我们可以采用迭代从基础值1,2累加上去。

代码1（递归）：

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

代码2（迭代）：

public class Solution {public int climbStairs(int n) {if(n==1||n==2){return n;}int n1=1;int n2=2;for(int i=3;i<=n;i++){int temp=n1+n2;n1=n2;n2=temp;}return n2;}
}

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