Leetcode: Ugly Number II

最新推荐文章于 2021-06-08 16:10:03 发布

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最新推荐文章于 2021-06-08 16:10:03 发布

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原文链接：http://www.cnblogs.com/EdwardLiu/p/5068745.html

本文介绍了一种高效算法用于查找第n个丑数，即只包含质因数2、3、5的正整数。通过维护三个指针并依次乘以这三个数中的最小值，算法确保了生成的序列有序且仅包含所需类型的丑数。

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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.

Hint:

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

参 Lintcode: Ugly Number

注意12-14行千万不要写成else if，因为如果同时两个Pointer比如 twoPointer and threePointer 都满足得到最小，两个pointer都要+1

 1 public class Solution {
 2     public int nthUglyNumber(int n) {
 3         ArrayList<Integer> res = new ArrayList<Integer>();
 4         if (n <= 0) return Integer.MIN_VALUE;
 5         res.add(1);
 6         int twoPointer = 0;
 7         int threePointer = 0;
 8         int fivePointer = 0;
 9         for (int i=2; i<=n; i++) {
10             int min = Math.min(Math.min(res.get(twoPointer)*2, res.get(threePointer)*3), res.get(fivePointer)*5);
11             res.add(min);
12             if (min/res.get(twoPointer) == 2) twoPointer++;
13             if (min/res.get(threePointer) == 3) threePointer++;
14             if (min/res.get(fivePointer) == 5) fivePointer++;
15         }
16         return res.get(n-1);
17     }
18 }