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Write a program to find the nth super ugly number.
Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4.
Note:
(1) 1 is a super ugly number for any given primes.
(2) The given numbers in primes are in ascending order.
(3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
Analysis:
Similar to Ugly Number II, but maintain a heap + a index map to perform O(log(k)) look up of minimum values.
NOTE: While theoretically the complexity is O(Nlog(k)), in reality, it may be slower than a O(NK) solution using primitive data structure like in[], because this solution use high-level data structures like PQ and HashMap.
Solution:
public class Solution { public int nthSuperUglyNumber(int n, int[] primes) { if (n==1) return 1; if (primes.length==0) return -1; int[] nums = new int[n]; nums[0] = 1; int[] pList = new int[primes.length]; Arrays.fill(pList,0); PriorityQueue<Integer> values = new PriorityQueue<Integer>(); HashMap<Integer,List<Integer>> valueMap = new HashMap<Integer,List<Integer>>(); for (int i=0;i<pList.length;i++){ addValue(primes,nums,values,valueMap,pList,i); } for (int i=1;i<n;i++){ // Get the min value int minVal = values.poll(); nums[i] = minVal; List<Integer> pointers = valueMap.get(minVal); valueMap.remove(minVal); for (int p : pointers){ pList[p]++; addValue(primes,nums,values,valueMap,pList,p); } } return nums[n-1]; } public void addValue(int[] primes, int[] nums, PriorityQueue<Integer> values, HashMap<Integer,List<Integer>> valueMap, int[] pList, int index){ int val = nums[pList[index]]*primes[index]; if (!valueMap.containsKey(val)){ valueMap.put(val,new ArrayList<Integer>()); values.add(val); } valueMap.get(val).add(index); } }
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原文地址：http://www.cnblogs.com/lishiblog/p/5867040.html
