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  • LeetCode-Super Ugly Number

    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:https://www.cnblogs.com/lishiblog/p/5867040.html
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