Question
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
isUglyfor 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).
Solution -- Queue
根据hint，我们维护三个队列。分别用来存当前的ugly number与2，3，5相乘后的结果。
选三个队列的队头元素最小的那个即是下一个ugly number。
1 public class Solution { 2 public int nthUglyNumber(int n) { 3 if (n < 2) { 4 return 1; 5 } 6 Queue<Integer> twoNumbers = new LinkedList<Integer>(); 7 Queue<Integer> threeNumbers = new LinkedList<Integer>(); 8 Queue<Integer> fiveNumbers = new LinkedList<Integer>(); 9 twoNumbers.offer(2); 10 threeNumbers.offer(3); 11 fiveNumbers.offer(5); 12 int result = 1; 13 while (n > 1) { 14 int candidate1 = twoNumbers.peek(), candidate2 = threeNumbers.peek(), candidate3 = fiveNumbers.peek(); 15 // find minimum number 16 result = Math.min(candidate1, candidate2); 17 result = Math.min(candidate3, result); 18 // add new numbers 19 twoNumbers.offer(result * 2); 20 threeNumbers.offer(result * 3); 21 fiveNumbers.offer(result * 5); 22 // remove existing minimums 23 if (result == candidate1) { 24 twoNumbers.poll(); 25 } 26 if (result == candidate2) { 27 threeNumbers.poll(); 28 } 29 if (result == candidate3) { 30 fiveNumbers.poll(); 31 } 32 n--; 33 } 34 return result; 35 } 36 }