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 <= 5000 <= arr[i] <= 10^4
Solution: Sorting
Time complexity: O(nlogn)
Space complexity: O(1)
C++
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
// Author: Huahua class Solution { public: vector<int> sortByBits(vector<int>& arr) { sort(begin(arr), end(arr), [](const int a, const int b){ int key_a = __builtin_popcount(a); int key_b = __builtin_popcount(b); if (key_a != key_b) return key_a < key_b; return a < b; }); return arr; } }; |
Python3
|
1 2 3 4 |
# Author: Huahua class Solution: def sortByBits(self, arr: List[int]) -> List[int]: return sorted(arr, key=lambda x: (bin(x).count('1') << 16) + x) |
请尊重作者的劳动成果，转载请注明出处！花花保留对文章／视频的所有权利。
如果您喜欢这篇文章／视频，欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Paypal
Venmo
huahualeetcode
huahualeetcode
微信打赏
Be First to Comment