1356. Sort Integers by The Number of 1 Bits

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

这题也比较简单，排序条件是按1bit位的个数。快速计算bit1的个数，之前也说过，直接可以用

n & (n - 1)

来计算，它可以快速的消除现有的1bit位。

class SortIntegersByTheNumberOf1Bits : public Solution {
public:
    void Exec() {

    }
    static int countBits(int num) {
        int count = 0;
        while (num != 0) {
            num = num & (num - 1);
            count++;
        }
        return count;
    }
    vector<int> sortByBits(vector<int>& arr) {
        std::sort(arr.begin(), arr.end(), [](int a, int b) {
            int bitA = countBits(a);
            int bitB = countBits(b);
            if (bitA != bitB) {
                return bitA < bitB;
            }
            return a < b;
        });
        return arr;
    }
};