Leetcode 650. 2 Keys Keyboard 2指键盘 解题报告

这道题可以转化成，给一个数字N，初始K=1，C=0然后只允许你有两种操作：
1、K = K + C （paste）
2 、C = K （copy all）
问，如何操作可以使得最快的得到N

N>1时，其实这道题就是将N分解为M个数字的乘积，且M个数字的和最小。

比如：
2 = 1 * 1 = 2
3 = 1 * 1 *1 = 3
4 = 2 * 2 = 1* 1* 1 *1 =4
等等
那么最快的讲一个数分解为N个质数的和怎们办呢

Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:Copy All: You can copy all the characters present on the notepad (partial copy is not allowed).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: 3Output: 3Explanation: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:The n will be in the range [1, 1000].

给了两个方法，第一个是看别人的写的，第二个是我自己做的，速度似乎差不多

第一个主要是从小到大的去试探，尽量用小的数字去除就可以

大神简洁解法：39ms

class Solution(object):    def minSteps(self, n):        """        :type n: int        :rtype: int        """        res = 0        for i in range(2, n+1):            while (n % i == 0):                res += i                n /= i        return res

DP 递归 版 36ms

首先转化为两个数字的子问题，且A=B*C 且 B+C最小，就是指这两个数字的和最小（其实就是从B=sqrt(A),不断向下试探）找，然后对于这两个数字再进行递归。。记得加cache（不然就用字典，提前好到所有1000内的质数，这个更快）：

class Solution(object):    import math    def helper(self, n):        """        :type n: int        :rtype: int        """        if n in self.cache:            return self.cache[n]        for i in range(int(math.sqrt(n)),1,-1):            if n % i == 0:                a = self.helper(i)                b = self.helper(n / i)                self.cache[n] = a + b                return a+ b        self.cache[n] = n        return n    def minSteps(self, n):        """        :type n: int        :rtype: int        """        self.cache = dict()        self.cache[1] = 1        if n == 1:            return 0        return self.helper(n)