264. Ugly Number II

最新推荐文章于 2025-09-05 16:34:31 发布

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最新推荐文章于 2025-09-05 16:34:31 发布

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文章标签： java runtime

原文链接：https://segmentfault.com/a/1190000018344160

本文介绍了一种高效算法，用于查找序列中的第N个丑数。丑数定义为仅包含2、3、5作为质因数的正整数。通过使用三个指针分别跟踪2、3、5倍数的生成过程，算法能够在O(n)时间内找到目标数值。

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.

难度：medium

题目：写程序找出第n个丑数。丑数公因子只由2,3,5构正的正整数。

思路：三路指针。

Runtime: 2 ms, faster than 99.79% of Java online submissions for Ugly Number II.
Memory Usage: 36.8 MB, less than 32.39% of Java online submissions for Ugly Number II.

class Solution {
    public int nthUglyNumber(int n) {
        int[] ugly = new int[n];
        ugly[0] = 1;
        int p2 = 0, p3 = 0, p5 = 0;
        for (int i = 1; i < n; i++) {
            ugly[i] = Math.min(ugly[p2] * 2, Math.min(ugly[p3] * 3, ugly[p5] * 5));
            if (ugly[i] == ugly[p2] * 2) {
                p2++;
            }
            if (ugly[i] == ugly[p3] * 3) {
                p3++;
            }
            if (ugly[i] == ugly[p5] * 5) {
                p5++;
            }
        }
        
        return ugly[n - 1];
    }
}