LeetCode 4 Keys Keyboard

原题链接在这里：https://leetcode.com/problems/4-keys-keyboard/description/

题目：

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

Note:

1. 1 <= N <= 50
2. Answers will be in the range of 32-bit signed integer.

题解：

DP问题.

初始化, dp[i] = i. i在[1,N]就是直接输入'A'.

状态转移, dp[i] = Math.max(dp[i], (i-j-1)*dp[j]), j在[1,i-3]区间内.  Ctrl A, Ctrl C, Ctrl V用了3个操作，所以j是到i-3. 本来已经有了d[j]个'A', 三个操作做完又加了dp[j]个'A', 还剩下i-3-j个操作，剩下的每一次都加dp[j]个'A'. 总共dp[j] + dp[j] + (i-3-j)*dp[j] = (i-j-1)*dp[j].

答案dp[N].

Time Complexity: O(N^2).

Space: O(N).

AC Java:

 class Solution {
     public int maxA(int N) {
         int [] dp = new int[N+1];
         for(int i = 1; i<=N; i++){
             dp[i] = i;
             for(int j = 1; j<=i-3; j++){
                 dp[i] = Math.max(dp[i], (i-j-1)*dp[j]);
             }
         }
         return dp[N];
     }
 }

类似2 Keys Keyboard.