题解 | #跳台阶#

public class Solution  {
    //公式:f(n)= f(n-1)+f(n-2);
    public  int jumpFloor(int number) {
       if(number == 1){
        return 1;
       }
        //a 初始值为1
        int a = 1;
         //a 初始值为2
        int b = 2;
        //临时变量
        int c = 0;
        for (int i = 2; i < number; i++) {
            //用c 临时储存f（n-1）+f(n-2)
            c = a + b;
            //a 前一轮b的值
            a = b;
            //b=存f（n-1）+f(n-2)
            b = c;
        }
         return b;
    }

}