(LeetCode 264) Ugly Number II

Q:
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).

如果一个数的因子只有2,3,5那么就说它是一个丑数，特殊的（1也是丑数），求第n个丑数。

solution：
很惭愧，这道题想了很久没有想出来，最后还是参考的别人的解法。
代码不难，关键是很灵巧，比较难想到。
代码不难，但是讲起来很晦涩，所以我就不讲解了，大家按照代码跑两个循环就能懂其中的机制了。。。

class Solution {public:    int nthUglyNumber(int n) {        int *uglyNums = new int[n];        int index_2=0;        int index_3=0;        int index_5=0;        uglyNums[0]=1;        for(int i=1;i<n;i++){            int min = mins( uglyNums[index_2]*2,                            uglyNums[index_3]*3,                            uglyNums[index_5]*5);            if(min==uglyNums[index_2]*2)index_2++;            if(min==uglyNums[index_3]*3)index_3++;            if(min==uglyNums[index_5]*5)index_5++;            uglyNums[i]=min;        }        return uglyNums[n-1];    }    int mins(int x,int y,int z){        int m = (x<y)?x:y;        return (m<z)?m:z;    }};

Reference Link:
http://my.oschina.net/Tsybius2014/blog/495962