1. 把历史丑数都存下来，每次*2或者*3或者*5，找到最小的那个丑数
2. 加到丑数结果集
3. 复杂度是n*2

1. leetcode的测试例子比较多，res之前的元素类型是int，但是超出限制了，改成long long，leetcode点也过啦

class Solution {
public:
    vector<long long> res;
    int GetUglyNumber_Solution(int index) {
        if (index == 0)
            return 0;
        res.push_back(1);
        int minUglyNumber;
        vector<int> factors;
        factors.push_back(2);
        factors.push_back(3);
        factors.push_back(5);
        while(index > 1) {
            minUglyNumber = -1;
            for (int i = 0; i < res.size(); i++) {
                for (int j = 0; j < factors.size(); j++) {
                    if (res[i]*factors[j] <= res[res.size()-1])
                        continue;
                    if (minUglyNumber == -1 || res[i]*factors[j] < minUglyNumber)
                        minUglyNumber = res[i]*factors[j];
                }
            }
            res.push_back(minUglyNumber);
            /*for (int i = 0; i < res.size(); i++)
                cout << res[i] << " ";
            cout << endl;
            while (res.size() > 9) {
                res.erase(res.begin(),res.begin()+1);
            }*/
            index--;
        }
        return res[res.size()-1];
    }
};