- 设铺第n个砖块可能的方法数为f(n)
- 铺每个砖块可以横着铺(占两格,f(n-2)),也可以竖着铺(只占一格,f(n-1)),即f(n) = f(n-1) + f(n-2)
public class Solution { public int RectCover(int target) { // 排除初始情况 if(target <= 1) return target; // 给出初始数据进行计算 int first = 1, second = 2; while(target-- > 2) { second += first; first = second - first; } return second; } }