[Leetcode] 625. Minimum Factorization 解题报告

题目：

Given a positive integer a, find the smallest positive integer b whose multiplication of each digit equals to a.

If there is no answer or the answer is not fit in 32-bit signed integer, then return 0.

Example 1
Input:

Output:

Example 2
Input:

Output:

思路：

可以观察到两个事实：1）n位数一定小于n+1位的数；2）如果几个结果都是n位数，那么将各个位上的数按照升序排列，则形成的n位数结果是最小的。所以我们的思路就是尽量让结果的位数小，并且让因数从大到小。所以可以让i从9循环到2，按照顺序找出a中所有9-2之间的因数（注意不是质因数），然后将它们按照升序排列，形成的结果即为题目所求。我们这里让d < 10是为了防止int溢出。

代码：

class Solution {
public:int smallestFactorization(int a) {if (a < 2) {return a;}long l = 0;for (int i = 9, d = 0; i >= 2 && d < 10; --i) {while (a % i == 0 && d < 10) {l += i * pow(10, d++);a /= i;}}return a > 1 || l > INT_MAX ? 0 : l;}
};

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