leetcode 264 ugly Number 2

Premium ugly number! In the end it is the first of several ugly?

This question is based on the original number of ugly
lookingThe first fewNumber ugly

Failed attempts

Analyzing directly hit the table
all numbers in ugly 1690
Sure timeout benefits serving

No one judge
would need to find data rule the

Observation available
ugly excellent base number 2, 3, 5 is the number of still ugly - Simplify Analyzing

So ugly number may consist of the following list

2	1* 2 2 * 2 3 * 2 4 * 2
3	1* 3 2 * 3 3 * 3 4 * 3
5	1* 2 2 * 2 3 * 2 4 * 2

practice:

• Maintenance three list - ugly2, ugly3, ugly5
  respectively store ** now * 2, now * 3 , now * 5 **
• Take three at the beginning of the list Minimum- The number cnt number of ugly (the equivalent logarithmic ugly sorting process)
• Until cnt reaches the n - that is to take the number of the n-th ugly

code show as below:

class Solution {
public:
    int nthUglyNumber(int n) {
        list<long long> ugly2;
        list<long long> ugly3;
        list<long long> ugly5;
        long long now = 1;
        int cnt = 1;
        while (cnt != n) {
            ugly2.push_back(now * 2);
            ugly3.push_back(now * 3);
            ugly5.push_back(now * 5);

            long long min2 = ugly2.front();
            long long min3 = ugly3.front();
            long long min5 = ugly5.front();

            long long MIN = min(min2, min(min3, min5));
            if (min2 == MIN) ugly2.pop_front();
            if (min3 == MIN) ugly3.pop_front();
            if (min5 == MIN) ugly5.pop_front();
            now = MIN;
            cnt++;
        }
        return now;
    }
};