【leetcode】1636. Sort Array by Increasing Frequency

题目如下:

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

解题思路:先统计每个值出现的次数,然后排序即可。

代码如下:

class Solution(object):
    def frequencySort(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        dic = {}
        for i in nums:
            dic[i] = dic.setdefault(i,0)+1
        pair = []
        for key,val in dic.iteritems():
            pair.append((key,val))

        def cmpf(p1,p2):
            if p1[1] != p2[1]:
                return p1[1] - p2[1]
            return p2[0] - p1[0]

        pair.sort(cmp=cmpf)

        res = []

        for (key,val) in pair:
            res += [key] * val

        return res
原文地址:https://www.cnblogs.com/seyjs/p/14931266.html