[Leetcode] 651. 4 Keys Keyboard 解题报告

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.

思路：

1、动态规划：

和上一道题目比较类似：我们要生成n个‘A’可以有两种途径：1）在键盘上敲击n次‘A’；2）采用i步生成maxA(i)个‘A’，然后用n - i步执行1次Ctrl A，1次Ctrl C，以及n - i- 2次Ctrl V，这样最终就可以生成n - i - 1个maxA(i)。由于n - i - 2 >= 1，所以i <= n - 3。如果我们定义dp[i]表示i步可以生成的‘A’的最大个数，则可以方便地实现基于动态规划的版本。

2、递推法：

O(1)的时间复杂度和空间复杂度，不过我是真的没有看懂，给个链接吧：O(1) time O(1) space c++ solution, possibly shortest and fastest。

代码：

1、动态规划：

class Solution {
public:
    int maxA(int N) {
        vector<int> dp(N + 1, 0);
        for (int i = 0; i <= N; ++i) {
            dp[i] = i;      // simply print 'A' i times
            for (int j = 1; j <= i - 3; ++j) {
                dp[i] = max(dp[i], dp[j] * (i - j - 1));
            }
        }
        return dp[N];
    }
};

2、递推法：

class Solution {
public:
    int maxA(int N) {
        if (N <= 6) {
            return N;
        }
        if (N == 10) {
            return 20;
        }
        int n = N / 5 + 1, n3 = n * 5 - 1 - N;
        return pow(3, n3) * pow(4, n - n3);
    }
};