[Leetcode] #263#264 Ugly Number I & II

奔跑の小乌龟

于 2017-05-07 13:12:05 发布

阅读量360

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/qq_34264472/article/details/71320502

LeetCode 专栏收录该内容

117 篇文章

订阅专栏

本文介绍了一种用于判断及寻找Ugly Number的高效算法。Ugly Number定义为只包含质因数2、3和5的正整数。文章提供了两个主要的解决方案：一是检查一个给定的数字是否为Ugly Number；二是找到第n个Ugly Number。通过使用动态规划的方法，算法能够有效地解决这两个问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Discription

Write a program to check whether a given number is an ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while14 is not ugly since it includes another prime factor 7.

Note that 1 is typically treated as an ugly number.

Solution

bool isUgly(int num) {
	while (num % 2 == 0 && num)
		num /= 2;
	while (num % 3 == 0 && num)
		num /= 3;
	while (num % 5 == 0 && num)
		num /= 5;
	return num == 1;
}

Discription

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, and n does not exceed 1690.

Solution

int nthUglyNumber(int n) {
	vector<int> res(n, 1);
	int index2 = 0, index3 = 0, index5 = 0;
	for (int i = 1; i<n; i++) {
		res[i] = min(min(res[index2] * 2, res[index3] * 3), res[index5] * 5);
		if (res[i] == res[index2] * 2) index2++;
		if (res[i] == res[index3] * 3) index3++;
		if (res[i] == res[index5] * 5) index5++;
	}
	return res[n - 1];
}

GitHub-Leetcode: https://github.com/wenwu313/LeetCode