[LeetCode] 264. Ugly Number II 丑陋數之二

 

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

Hint:

1. The naive approach is to call isUgly for every number until you reach the n^th one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones.
2. An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
3. 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.
4. Assume you have U_k, the k^th ugly number. Then U_k+1 must be Min(L₁ * 2, L₂ * 3, L₃ * 5).

 

這道題是之前那道 Ugly Number 的拓展，這里讓找到第n個丑陋數，還好題目中給了很多提示，基本上相當於告訴我們解法了，根據提示中的信息，丑陋數序列可以拆分為下面3個子列表：

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

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

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

仔細觀察上述三個列表，可以發現每個子列表都是一個丑陋數分別乘以 2，3，5，而要求的丑陋數就是從已經生成的序列中取出來的，每次都從三個列表中取出當前最小的那個加入序列，請參見代碼如下：

 

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();
    }
};

 

我們也可以使用最小堆來做，首先放進去一個1，然后從1遍歷到n，每次取出堆頂元素，為了確保沒有重復數字，進行一次 while 循環，將此時和堆頂元素相同的都取出來，然后分別將這個取出的數字乘以 2，3，5，並分別加入最小堆。這樣最終 for 循環退出后，堆頂元素就是所求的第n個丑陋數，參見代碼如下：

 

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：

https://github.com/grandyang/leetcode/issues/264

 

類似題目：

Super Ugly Number

Ugly Number

Happy Number

Count Primes

Merge k Sorted Lists

Perfect Squares

 

參考資料：

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

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

https://leetcode.com/problems/ugly-number-ii/discuss/69364/My-16ms-C%2B%2B-DP-solution-with-short-explanation

https://leetcode.com/problems/ugly-number-ii/discuss/69368/Elegant-C%2B%2B-Solution-O(N)-space-time-with-detailed-explanation.

 

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