264. Ugly Number II

O(n). index_x保存以上一次增长之前的数字位置，factor_x记录当前最小的可能

    public int nthUglyNumber(int n) {
        if(n <= 0) {
            return 0;
        }
        int idx2 = 0, idx3 = 0, idx5 = 0;
        int factor2 = 2, factor3 = 3, factor5 = 5;
        int[] ugly = new int[n];
        ugly[0] = 1;
        for(int i = 1; i < n; i++) {
            ugly[i] = Math.min(factor2, Math.min(factor3, factor5));
            if(ugly[i] == factor2) {
                factor2 = ugly[++idx2] * 2;
            }
            if(ugly[i] == factor3) {
                factor3 = ugly[++idx3] * 3;
            }
            if(ugly[i] == factor5) {
                factor5 = ugly[++idx5] * 5;
            }
        }
        return ugly[n-1];
    }