LeetCode Ugly Number || DP
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思路:
参考DISCUSS:https://leetcode.com/discuss/52716/o-n-java-solution
每个丑数都仅由2,3,5质因子构成,因此通过这3个数不断的乘以2,3,5就会产生所有其余的丑数,(每个丑数乘以2,3,5会得到一组新丑数)关键是顺序的问题。
时间复杂度O(N),空间复杂度O(N)。
java code:
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]; }}
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