[LeetCode] 650. 2 Keys Keyboard 两键的键盘

Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:

1. Copy All: You can copy all the characters present on the notepad (partial copy is not allowed).
2. Paste: You can paste the characters which are copied last time.

Given a number n. You have to get exactly n 'A' on the notepad by performing the minimum number of steps permitted. Output the minimum number of steps to get n 'A'.

Example 1:

Input: 3
Output: 3
Explanation:
Intitally, we have one character 'A'.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get 'AA'.
In step 3, we use Paste operation to get 'AAA'.

Note:

1. The n will be in the range [1, 1000].

初始化一个记事本，里面只含有一个字符'A'，每次可以执行一个操作，共有两个操作：拷贝全部，粘贴上一次的拷贝。给一个数组n，求要达到n个字符，最少需要操作几步。

技巧是循环找出n的分解因子i，这个因子就是可以当作模块的个数，再算出模块的长度n/i，调用递归，加上模块的个数i来更新结果res

解法1：递归。

解法2：DP

Java:

class Solution {
    public int minSteps(int n) {
        int[] dp = new int[n+1];

        for (int i = 2; i <= n; i++) {
            dp[i] = i;
            for (int j = i-1; j > 1; j--) {
                if (i % j == 0) {
                    dp[i] = dp[j] + (i/j);
                    break;
                }
                
            }
        }
        return dp[n];
    }
}　　

Python:

class Solution(object):
    def minSteps(self, n):
        """
        :type n: int
        :rtype: int
        """
        result = 0
        p = 2
        # the answer is the sum of prime factors
        while p**2 <= n:
            while n % p == 0:
                result += p
                n //= p
            p += 1
        if n > 1:
            result += n
        return result　　

C++:

class Solution {
public:
    int minSteps(int n) {
        if (n == 1) return 0;
        int res = n;
        for (int i = n - 1; i > 1; --i) {
            if (n % i == 0) {
                res = min(res, minSteps(n / i) + i);
            }
        }
        return res;
    }
};

C++:

class Solution {
public:
    int minSteps(int n) {
        vector<int> dp(n + 1, 0);
        for (int i = 2; i <= n; ++i) {
            dp[i] = i;
            for (int j = i - 1; j > 1; --j) {
                if (i % j == 0) {
                    dp[i] = min(dp[i], dp[j] + i / j);
                }
            }
        }
        return dp[n];
    }
};

C++:

class Solution {
public:
    int minSteps(int n) {
        int res = 0;
        for (int i = 2; i <= n; ++i) {
            while (n % i == 0) {
                res += i;
                n /= i;
            }
        }
        return res;
    }
};

　　

　　

　　

 

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