LeetCode 264 Ugly Number II (递推)

最新推荐文章于 2025-12-28 17:52:30 发布

原创最新推荐文章于 2025-12-28 17:52:30 发布 · 645 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #丑数2

LeetCode 同时被 2 个专栏收录

334 篇文章

订阅专栏

Leetcode Medium

244 篇文章

订阅专栏

本文介绍了一种求解第N个丑数的有效算法。丑数是指只包含质因数2、3和5的正整数。通过动态规划的方法，使用最小值选择策略，确保了时间复杂度为O(n)。

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.

题目链接：https://leetcode.com/problems/ugly-number-ii/

题目分析：用一个uglyNum数组记录，保证按升序构造，用当前的每个丑数，分别乘2，3，5其中比最大的大的最少的就是当前的下一个丑数，递推时间复杂度O(n)

public class Solution {
    
    public int Min3(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }
    
    public int nthUglyNumber(int n) {
        int[] uglyNum = new int[n + 5];
        uglyNum[1] = 1;
        int i = 1, i2 = 1, i3 = 1, i5 = 1, cur;
        while(i <= n) {
            cur = Min3(uglyNum[i2] * 2, uglyNum[i3] * 3, uglyNum[i5] * 5);
            if(cur == uglyNum[i2] * 2) {
                i2 ++;
            }
            if(cur == uglyNum[i3] * 3) {
                i3 ++;
            }
            if(cur == uglyNum[i5] * 5) {
                i5 ++;
            }
            uglyNum[++ i] = cur;
        }
        return uglyNum[n];
    }
}