leetcode 264. Ugly Number II

Write a program to find the n-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note that 1 is typically treated as an ugly number.

class Solution {
	struct Node
	{
		int val;
		vector<int>bb;
		Node(int x) :val(x){
			bb.push_back(5);
			bb.push_back(3);
			bb.push_back(2);
		}
	};
public:
	int nthUglyNumber(int n) {
		clock_t start, finish;
		double Total_time;
		start = clock();
		if (n == 1)
			return 1;
		vector<int>cc;
		cc.push_back(1);
		vector<Node>candi;
		candi.push_back(Node(1));
		while (cc.size() != n)
		{
			int min = 2147483647;
			int ii = -1;
			int i = 0;
			while (candi.size() > i)
			{
				while (candi[i].val*candi[i].bb.back() <= cc.back())
					candi[i].bb.pop_back();
				if (candi[i].val*candi[i].bb.back() < min)
				{
					min = candi[i].val*candi[i].bb.back();
					ii = i;
				}
				i++;
				if (i < candi.size() && candi[i].bb.size() == 3 && candi[i].val * 2 >= min)
					break;
			}
			cc.push_back(min);
			candi.push_back(Node(min));
			candi[ii].bb.pop_back();
			if (candi[ii].bb.empty())
				candi.erase(candi.begin() + ii, candi.begin() + ii + 1);

		}
		finish = clock();
		Total_time = (double)(finish - start) / CLOCKS_PER_SEC;
		printf("%f seconds\n", Total_time);
		return cc.back();
	}
};

这个超时了，想了想性能的瓶颈可能是每次找min时计算太多，其实上次算出来的结果可以存下来，这样可能会减少一些计算。

改进之后accept

class Solution {
public:
	int nthUglyNumber(int n) {
		clock_t start, finish;
		double Total_time;
		start = clock();
		if (n == 1)
			return 1;
		vector<int>cc;
		cc.push_back(1);
		set<int>m2, m3, m5;
		m2.insert(1);
		m3.insert(1);
		m5.insert(1);
		while (cc.size() != n)
		{
			while (*m2.begin() * 2 <= cc.back())
				m2.erase(m2.begin());
			int a2 = *m2.begin() * 2;
			while (*m3.begin() * 3 <= cc.back())
				m3.erase(m3.begin());
			int a3 = *m3.begin() * 3;
			while (*m5.begin() * 5 <= cc.back())
				m5.erase(m5.begin());
			int a5 = *m5.begin() * 5;
			int min = a2 < a3 ? a2 : a3;
			min = min < a5 ? min : a5;
			if (min == a2)
				m2.erase(m2.begin());
			if (min==a3)
				m3.erase(m3.begin());
			if (min==a5)
				m5.erase(m5.begin());
			m2.insert(min);
			m3.insert(min);
			m5.insert(min);
			cc.push_back(min);
		}

		finish = clock();
		Total_time = (double)(finish - start) / CLOCKS_PER_SEC;
		printf("%f seconds\n", Total_time);
		return cc.back();
	}
};