[LeetCode] 264. Ugly Number II #264

grandyang · 2019-05-30T16:18:08Z

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

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.

Example:

Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note:

1 is typically treated as an ugly number.
n does not exceed 1690.

Hint:

The naive approach is to call isUgly for every number until you reach the nth one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones.
An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
The key is how to maintain the order of the ugly numbers. Try a similar approach of merging from three sorted lists: L1, L2, and L3.
Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).

这道题是之前那道 Ugly Number 的拓展，这里让找到第n个丑陋数，还好题目中给了很多提示，基本上相当于告诉我们解法了，根据提示中的信息，丑陋数序列可以拆分为下面3个子列表：

(1) 1x2 , 2x2, 2x2 , 3x2, 3x2 , 4x2 , 5x2...

(2) 1x3, 1x3 , 2x3, 2x3, 2x3 , 3x3, 3x3...

(3) 1x5, 1x5, 1x5, 1x5 , 2x5, 2x5, 2x5...

仔细观察上述三个列表，可以发现每个子列表都是一个丑陋数分别乘以 2，3，5，而要求的丑陋数就是从已经生成的序列中取出来的，每次都从三个列表中取出当前最小的那个加入序列，请参见代码如下：

解法一：

class Solution {
public:
    int nthUglyNumber(int n) {
        vector<int> res(1, 1);
        int i2 = 0, i3 = 0, i5 = 0;
        while (res.size() < n) {
            int m2 = res[i2] * 2, m3 = res[i3] * 3, m5 = res[i5] * 5;
            int mn = min(m2, min(m3, m5));
            if (mn == m2) ++i2;
            if (mn == m3) ++i3;
            if (mn == m5) ++i5;
            res.push_back(mn);
        }
        return res.back();
    }
};

我们也可以使用最小堆来做，首先放进去一个1，然后从1遍历到n，每次取出堆顶元素，为了确保没有重复数字，进行一次 while 循环，将此时和堆顶元素相同的都取出来，然后分别将这个取出的数字乘以 2，3，5，并分别加入最小堆。这样最终 for 循环退出后，堆顶元素就是所求的第n个丑陋数，参见代码如下：

解法二：

class Solution {
public:
    int nthUglyNumber(int n) {
        priority_queue<long, vector<long>, greater<long>> pq;
        pq.push(1);
        for (long i = 1; i < n; ++i) {
            long t = pq.top(); pq.pop();
            while (!pq.empty() && pq.top() == t) {
                t = pq.top(); pq.pop();
            }
            pq.push(t * 2);
            pq.push(t * 3);
            pq.push(t * 5);
        }
        return pq.top();
    }
};

Github 同步地址：

类似题目：

参考资料：

https://leetcode.com/problems/ugly-number-ii/

https://leetcode.com/problems/ugly-number-ii/discuss/69372/Java-solution-using-PriorityQueue