算法 中等 | 4. 丑数 II

题目描述

设计一个算法，找出只含素因子2，3，5 的第 n 小的数。
符合条件的数如：1, 2, 3, 4, 5, 6, 8, 9, 10, 12…

样例1

输入：9
输出：10

样例2

输入：1
输出：1

java题解

题解1

class Solution {
    /**
     * @param n an integer
     * @return the nth prime number as description.
     */
    public int nthUglyNumber(int n) {
        // Write your code here
        Queue<Long> Q = new PriorityQueue<Long>();
        HashSet<Long> inQ = new HashSet<Long>();
        Long[] primes = new Long[3];
        primes[0] = Long.valueOf(2);
        primes[1] = Long.valueOf(3);
        primes[2] = Long.valueOf(5);
        for (int i = 0; i < 3; i++) {
            Q.add(primes[i]);
            inQ.add(primes[i]);
        }
        Long number = Long.valueOf(1);
        for (int i = 1; i < n; i++) {
            number = Q.poll();
            for (int j = 0; j < 3; j++) {
                if (!inQ.contains(primes[j] * number)) {
                    Q.add(number * primes[j]);
                    inQ.add(number * primes[j]);
                }
            }
        }
        return number.intValue();
    }
};

题解2

class Solution {
    /**
     * @param n an integer
     * @return the nth prime number as description.
     */
    public int nthUglyNumber(int n) {
        List<Integer> uglys = new ArrayList<Integer>();
        uglys.add(1);
        
        int p2 = 0, p3 = 0, p5 = 0;
        // p2, p3 & p5 share the same queue: uglys

        for (int i = 1; i < n; i++) {
            int lastNumber = uglys.get(i - 1);
            while (uglys.get(p2) * 2 <= lastNumber) p2++;
            while (uglys.get(p3) * 3 <= lastNumber) p3++;
            while (uglys.get(p5) * 5 <= lastNumber) p5++;
            
            uglys.add(Math.min(
                Math.min(uglys.get(p2) * 2, uglys.get(p3) * 3),
                uglys.get(p5) * 5
            ));
        }

        return uglys.get(n - 1);
    }
};

C++题解

class Solution {
public:
    /*
     * @param n an integer
     * @return the nth prime number as description.
     */
    int nthUglyNumber(int n) {
        int *uglys = new int[n];
        uglys[0] = 1;
        int next = 1;
        int *p2 = uglys;
        int *p3 = uglys;
        int *p5 = uglys;
        while (next < n){
            int m = min(min(*p2 * 2, *p3 * 3), *p5 * 5);
            uglys[next] = m;
            while (*p2 * 2 <= uglys[next])
                *p2++;
            while (*p3 * 3 <= uglys[next])
                *p3++;
            while (*p5 * 5 <= uglys[next])
                *p5++;
            next++;
        }
        int uglyNum = uglys[n - 1];
        delete[] uglys;
        return uglyNum;
    }
};

python题解

import heapq
class Solution:
    """
    @param {int} n an integer.
    @return {int} the nth prime number as description.
    """
    def nthUglyNumber(self, n):
        heap = [1]
        visited = set([1])
        
        val = None
        for i in range(n):
            val = heapq.heappop(heap)
            for factor in [2, 3, 5]:
                if val * factor not in visited:
                    visited.add(val * factor)
                    heapq.heappush(heap, val * factor)
            
        return val