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

 1     public int nthUglyNumber(int n) {
 2         if(n <= 0) {
 3             return 0;
 4         }
 5         int idx2 = 0, idx3 = 0, idx5 = 0;
 6         int factor2 = 2, factor3 = 3, factor5 = 5;
 7         int[] ugly = new int[n];
 8         ugly[0] = 1;
 9         for(int i = 1; i < n; i++) {
10             ugly[i] = Math.min(factor2, Math.min(factor3, factor5));
11             if(ugly[i] == factor2) {
12                 factor2 = ugly[++idx2] * 2;
13             }
14             if(ugly[i] == factor3) {
15                 factor3 = ugly[++idx3] * 3;
16             }
17             if(ugly[i] == factor5) {
18                 factor5 = ugly[++idx5] * 5;
19             }
20         }
21         return ugly[n-1];
22     }

 

posted @ 2016-08-13 11:02  warmland  阅读(205)  评论(0)  编辑  收藏  举报