• 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;
    }
}