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?
这是一道动态规划的题目，每一步的状态都与之前一步的状态有关，这道题总结出的规律就是f(n)=f(n-1)+f(n-2)。这里最直观的想法就是使用递归：
var climbStairs = function(n) {
if (n===1) {
return 1;
} else if (n===2) {
return 2;
} else {
return climbStairs(n-1)+climbStairs(n-2);
}
};
但是递归里有太多重复的计算，所以会比较慢。那就使用最普通的方法一个一个加：
var climbStairs = function(n) {
if(n===0||n==1||n==2){
return n;
}
var step1 = 1;
var step2 = 2;
var stepn = 0;
var i = 3;
while(i<=n){
stepn = step1 + step2;
step1 = step2;
step2 = stepn;
i++;
}
return stepn;
};
