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?

熟悉斐波那契数列的一看就知道斐波那契数列是0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...

其通项是f[k] = f[k-1]+f[k-2]，k>=2

其中f[0] = 0, f[1] = 1;

而此题类似斐波那契数但初始值有点不同

f[0] = 0, f[1]= 1, f[2] = 2， 对其迭代就行

int climbStairs(int n){    if(n == 0 || n == 1 || n == 2) return n;    int stepOne = 1, stepTwo = 2, result = 0;    for(int i = 3; i <= n ; ++ i){        result = stepOne + stepTwo;        stepOne = stepTwo;        stepTwo = result;    }    return result;}