[leetcode-625-Minimum Factorization]

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48 

68

15

35

思路：

For a given n, following are the two cases to be considered.
Case 1: n < 10 When n is smaller than n, the output is always n+10. For example for n = 7, output is 17. For n = 9, output is 19.

Case 2: n >= 10 Find all factors of n which are between 2 and 9 (both inclusive). The idea is to start searching from 9 so that the number of digits in result are minimized. For example 9 is preferred over 33 and 8 is preferred over 24.
Store all found factors in an array. The array would contain digits in non-increasing order, so finally print the array in reverse order.

Following is the implementation of above concept.

 int smallestFactorization(int a) {
              if(a<10)return a;
      vector<int> tmp;
      for(int i=9;i>1;i--)
      {
    while(a%i==0)
    {
      a = a/ i;
      tmp.push_back(i);
    }
      }
      if(a>10)return  0;
       long long ret =0;
       for(int i =tmp.size()-1;i>=0;i--)
       {
     ret = ret*10+tmp[i];
     if(ret>2147483647) return 0;
      }
      
      return ret;
    }

参考：

http://www.geeksforgeeks.org/find-smallest-number-whose-digits-multiply-given-number-n/

[leetcode-625-Minimum Factorization]

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原文地址：http://www.cnblogs.com/hellowooorld/p/7044050.html