191. Number of 1 Bits
Given a positive integer n, write a function that returns the number of set bits in its binary representation (also known as the Hamming weight).
int hammingWeight(uint32_t n) {int count = 0;while (n) {count += n & 1; // 检查最低位是否为 1n >>= 1; // 右移一位}return count;
}int main() {uint32_t n = 11; // 二进制为 1011cout << "Number of set bits: " << hammingWeight(n) << endl; // 输出 3return 0;
}
1356. Sort Integers by The Number of 1 Bits
You are given an integer array arr. 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 array after sorting it.
给定一个整数数组 arr，要求按照以下规则对数组进行排序：
- 按二进制表示中 1 的个数升序排序。
- 如果两个数的二进制表示中 1 的个数相同，则按数值大小升序排序。
最后返回排序后的数组。
Easy Understanding Version
int countOnes(int num) {int count = 0;while (num) {count += num % 2;num = num / 2;}return count;
}bool wayToSort(int i, int j) {int countA = countOnes(i);int countB = countOnes(j);if (countA == countB) {return i < j;}return cntA < cntB;
}
vector<int> sortByBits(vector<int>& arr) {sort(arr.begin(), arr.end(), wayToSort);return arr;
}
Compact Version
int countOnes(int n) {return __builtin_popcount(n); // 计算二进制中 1 的个数
}vector<int> sortByBits(vector<int>& arr) {sort(arr.begin(), arr.end(), [](int a, int b) {int countA = countOnes(a);int countB = countOnes(b);if (countA == countB) {return a < b; // 如果 1 的个数相同，按数值升序排序}return countA < countB; // 否则按 1 的个数升序排序});return arr;
}