264. Ugly Number II

本文介绍了一种高效算法用于找出序列中的第N个丑数。丑数定义为仅包含2、3、5作为质因数的正整数。文章详细解释了如何使用三个索引维护丑数序列并确保其正确生成。

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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, and n does not exceed 1690.

Hint:

  1. 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.
  2. An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
  3. 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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

我们需要3个index来推进ugly序列。推进原则是uglies[i] = Math.min(uglies[t2] * 2, Math.min(uglies[t3] * 3, uglies[t5] * 5));

代码如下:

public class Solution {
    public int nthUglyNumber(int n) {
        int[] uglies = new int[n];
        int t2 = 0, t3 = 0, t5 = 0;
        uglies[0] = 1;
        for (int i = 1; i < n; i ++) {
            uglies[i] = Math.min(uglies[t2] * 2, Math.min(uglies[t3] * 3, uglies[t5] * 5));
            if (uglies[i] == uglies[t2] * 2) t2++;
            if (uglies[i] == uglies[t3] * 3) t3++;
            if (uglies[i] == uglies[t5] * 5) t5++;
        }
        return uglies[n - 1];
    }
}

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