第n个丑数

public class Solution {
    public int GetUglyNumber_Solution(int index) {
       if(index==0) return 0;//注意0的情况，没有加这一行牛客网上显示通过0%
       int count=1;
        int index_2=0,index_3=0,index_5=0;//三个因子的指针
        int number=1;
        int[] uglyNumber=new int[index];//用于保存前index个丑数
        uglyNumber[0]=1;
        while(count<index){
            int tmpnumber_2=uglyNumber[index_2]*2;
            int tmpnumber_3=uglyNumber[index_3]*3;
            int tmpnumber_5=uglyNumber[index_5]*5;
            number=Math.min(tmpnumber_2,Math.min(tmpnumber_3, tmpnumber_5));//最小的数作为下一个丑数
            uglyNumber[count]=number;
            while(uglyNumber[index_2]*2<=uglyNumber[count])
                index_2++;
            while(uglyNumber[index_3]*3<=uglyNumber[count])
                index_3++;
            while(uglyNumber[index_5]*5<=uglyNumber[count])
                index_5++;
            count++;
        }
        return number;
    }
}