递推套路
矩形覆盖
http://www.nowcoder.com/questionTerminal/72a5a919508a4251859fb2cfb987a0e6
跟跳台阶一样的解法，利用递推：
public class Solution {
public int rectCover(int target) {
if(target == 0)
return 0;
int a = 1, b = 1;
for(int i = 1; i < target; i++){
a = a + b;
b = a - b;
}
return a;
}
}
难怪小公司那么挑剔，让你们这些大佬把位置拿了