leetcode 264. Ugly Number II

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本文介绍了一种高效求解第N个丑数的方法。丑数是指只包含质因数2、3和5的正整数。文章提供了一个O(n)的Java解决方案，并通过将丑数序列划分为三个子序列来简化问题。

Total Accepted: 56978
Total Submissions: 176871
Difficulty: Medium
Contributor: LeetCode
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.

比较好懂的解法

O(n) Java solution
The idea of this solution is from this page:http://www.geeksforgeeks.org/ugly-numbers/

The ugly-number sequence is 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …
because every number can only be divided by 2, 3, 5, one way to look at the sequence is to split the sequence to three groups as below:

(1) 1×2, 2×2, 3×2, 4×2, 5×2, …
(2) 1×3, 2×3, 3×3, 4×3, 5×3, …
(3) 1×5, 2×5, 3×5, 4×5, 5×5, …
We can find that every subsequence is the ugly-sequence itself (1, 2, 3, 4, 5, …) multiply 2, 3, 5.

Then we use similar merge method as merge sort, to get every ugly number from the three subsequence.

Every step we choose the smallest one, and move one step after,including nums with same value.

Thanks for this author about this brilliant idea. Here is my java solution

public class Solution {
    public int nthUglyNumber(int n) {
        int[] ugly = new int[n];
        ugly[0] = 1;
        int index2 = 0, index3 = 0, index5 = 0;
        int factor2 = 2, factor3 = 3, factor5 = 5;
        for(int i=1;i<n;i++){
            int min = Math.min(Math.min(factor2,factor3),factor5);
            ugly[i] = min;
            if(factor2 == min)
                factor2 = 2*ugly[++index2];
            if(factor3 == min)
                factor3 = 3*ugly[++index3];
            if(factor5 == min)
                factor5 = 5*ugly[++index5];
        }
        return ugly[n-1];
    }
}