题解 | #跳台阶#

与斐波那契数一样，只是提前了一位而已。

class Solution {
public:
    int jumpFloor(int number) {
        if (number == 1) return 1;
        if (number == 2) return 2;
        int n1 = 1, n2 = 2, n3;
        for (int i = 3; i <= number; i++) {
            n3 = n1 + n2;
            n1 = n2;
            n2 = n3;
        }
        return n3;
    }
};