#题解# 跳台阶扩展问题
跳台阶扩展问题
https://www.nowcoder.com/practice/22243d016f6b47f2a6928b4313c85387?tpId=196&rp=1&ru=%2Fexam%2Foj&qru=%2Fexam%2Foj&sourceUrl=%2Fexam%2Foj%3Fpage%3D1%26pageSize%3D50%26search%3D%26tab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D196&difficulty=&judgeStatus=&tags=591&title=&gioEnter=menu
熟悉一下动态规划，从入门做起
递归自上而下，动态规划自下而上
class Solution {
public:
// 扩展版的跳台阶，其当前台阶可以由所有前面的台阶跳上来，而不局限于前两个台阶
// 所以需要记录前面所有的台阶数，当前台阶等于之前所有台阶数之和
int jumpFloorII(int number) {
int res = 1;
int sum = 2;
for (int i = 2; i <= number; ++i) {
res = sum;
sum += res;
}
return res;
}
};
