手写代码：青蛙跳台阶

//这个题采用动态规划的思想，它每次只能跳一阶或者两阶台阶；那么它跳到第n阶台阶就有两种情况，从第n-1阶台阶一次跳一阶；或者从n-2阶台阶一次跳两阶；那么依次类推，只要保留跳到n-1和n-2的情况就可以算出跳到n的次数；类似于斐波那契数列

class Solution {
    public:

        int jumpFloor(int number) {
              if (number <= 0)
                   return 0;
              if (number == 1)
                   return 1;
              if (number == 2)
                   return 2;
              int a = 1;
              int b = 2;
              int temp = 0;
              for (int i = 3; i <= number; i++)
              {
                   temp = a + b;
                   a = b;
                   b = temp;
              }
          return temp;
        }
};