LeetCode 1636. Sort Array by Increasing Frequency
原题链接在这里:https://leetcode.com/problems/sort-array-by-increasing-frequency/
题目:
Given an array of integers nums
, sort the array in increasing order based on the frequency of the values. If multiple values have the same frequency, sort them in decreasing order.
Return the sorted array.
Example 1:
Input: nums = [1,1,2,2,2,3] Output: [3,1,1,2,2,2] Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 3.
Example 2:
Input: nums = [2,3,1,3,2] Output: [1,3,3,2,2] Explanation: '2' and '3' both have a frequency of 2, so they are sorted in decreasing order.
Example 3:
Input: nums = [-1,1,-6,4,5,-6,1,4,1] Output: [5,-1,4,4,-6,-6,1,1,1]
Constraints:
1 <= nums.length <= 100
-100 <= nums[i] <= 100
题解:
HashMap to record num and its frequency.
Use bucket sort to add num into it corresponding frequency bucket.
Go through buckets and add value one by one.
Time Complexity: O(nlogk). n = nums.length. k is average size of bucket list item.
Space: O(n).
AC Java:
1 class Solution { 2 public int[] frequencySort(int[] nums) { 3 if(nums == null || nums.length == 0){ 4 return nums; 5 } 6 7 int n = nums.length; 8 int [] res = new int[n]; 9 HashMap<Integer, Integer> hm = new HashMap<>(); 10 for(int num : nums){ 11 hm.put(num, hm.getOrDefault(num, 0) + 1); 12 } 13 14 ArrayList<Integer>[] buckets = new ArrayList[n + 1]; 15 for(Map.Entry<Integer, Integer> entry : hm.entrySet()){ 16 int freq = entry.getValue(); 17 if(buckets[freq] == null){ 18 buckets[freq] = new ArrayList<Integer>(); 19 } 20 21 for(int i = 0; i < freq; i++){ 22 buckets[freq].add(entry.getKey()); 23 } 24 } 25 26 int count = 0; 27 for(int i = 0; i <= n; i++){ 28 if(buckets[i] != null){ 29 Collections.sort(buckets[i], (a, b) -> b - a); 30 for(int num : buckets[i]){ 31 res[count++] = num; 32 } 33 } 34 } 35 36 return res; 37 } 38 }
AC C++:
1 class Solution { 2 public: 3 vector<int> frequencySort(vector<int>& nums) { 4 unordered_map<int, int> map; 5 for(int num : nums){ 6 map[num]++; 7 } 8 9 sort(begin(nums), end(nums), [&](int a, int b){ 10 return map[a] == map[b] ? a > b : map[a] < map[b]; 11 }); 12 13 return nums; 14 } 15 };
AC Python:
1 class Solution: 2 def frequencySort(self, nums: List[int]) -> List[int]: 3 count = collections.Counter(nums) 4 return sorted(nums, key = lambda x: (count[x], -x))
类似Top K Frequent Elements, Sort Characters By Frequency, Top K Frequent Words.