[LeetCode]#1356. Sort Integers by The Number of 1 Bits

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]
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.
Input: arr = [10000,10000]
Output: [10000,10000]
Input: arr = [2,3,5,7,11,13,17,19]
Output: [2,3,5,17,7,11,13,19]
Input: arr = [10,100,1000,10000]
Output: [10,100,10000,1000]
  1. Convert arr to bit format.
  2. count(‘1') in each arr
  3. sort it by step 2 red mark value
  4. return blue mark value
class Solution:
def sortByBits(self, arr):
for i in sorted(arr):
sd = sorted(d, key=lambda i: i[1])ans=[]
for i in sd:
return ans
  1. https://leetcode.com/problems/sort-integers-by-the-number-of-1-bits/discuss/991454/Python%3A-99-Faster-Using-Tuples
  2. https://blog.csdn.net/u010758410/article/details/79737498




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