[抄题]:
Imagine you have a special keyboard with the following keys:
Key 1: (A)
: Print one 'A' on screen.
Key 2: (Ctrl-A)
: Select the whole screen.
Key 3: (Ctrl-C)
: Copy selection to buffer.
Key 4: (Ctrl-V)
: Print buffer on screen appending it after what has already been printed.
Now, you can only press the keyboard for N times (with the above four keys), find out the maximum numbers of 'A' you can print on screen.
Example 1:
Input: N = 3
Output: 3
Explanation:
We can at most get 3 A's on screen by pressing following key sequence:
A, A, A
Example 2:
Input: N = 7
Output: 9
Explanation:
We can at most get 9 A's on screen by pressing following key sequence:
A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
贪心算法可以从除法 and 乘法的角度来思考
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
因为要做全选、拷贝、粘贴:三个数以内要进行相乘
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
复制粘贴不是乘法就是除法
[复杂度]:Time complexity: O(n) Space complexity: O(1)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
class Solution { public int maxA(int N) { int max = N; for (int i = 1; i <= N - 3; i++) { max = Math.max(max, maxA(i) * (N - i - 1)); } return max; } }