【leetcode】1356. Sort Integers by The Number of 1 Bits

题目如下：

Given an integer array arr. You have to sort the integers in the array in ascending order by the number of 1‘s in their binary representation and in case of two or more integers have the same number of 1‘s you have to sort them in ascending order.

Return the sorted array.

Example 1:

Input: arr = [0,1,2,3,4,5,6,7,8]
Output: [0,1,2,4,8,3,5,6,7]
Explantion: [0] is the only integer with 0 bits.
[1,2,4,8] all have 1 bit.
[3,5,6] have 2 bits.
[7] has 3 bits.
The sorted array by bits is [0,1,2,4,8,3,5,6,7]

Example 2:

Input: arr = [1024,512,256,128,64,32,16,8,4,2,1]
Output: [1,2,4,8,16,32,64,128,256,512,1024]
Explantion: All integers have 1 bit in the binary representation, you should just sort them in ascending order.

Example 3:

Input: arr = [10000,10000]
Output: [10000,10000]

Example 4:

Input: arr = [2,3,5,7,11,13,17,19]
Output: [2,3,5,17,7,11,13,19]

Example 5:

Input: arr = [10,100,1000,10000]
Output: [10,100,10000,1000]

Constraints:

• 1 <= arr.length <= 500
• 0 <= arr[i] <= 10^4

解题思路：转换成二进制比较排序。

代码如下：

class Solution(object):
    def sortByBits(self, arr):
        """
        :type arr: List[int]
        :rtype: List[int]
        """
        bin_arr = []
        for i in arr:
            bin_arr.append((i,bin(i).count(‘1‘)))
        def cmpf(v1,v2):
            if v1[1] != v2[1]:
                return v1[1] - v2[1]
            return v1[0] - v2[0]
        bin_arr.sort(cmp=cmpf)
        res = []
        for (i1,i2) in bin_arr:
            res.append(i1)
        return res

【leetcode】1356. Sort Integers by The Number of 1 Bits

原文：https://www.cnblogs.com/seyjs/p/12349894.html