LeetCode 338.比特位计数

各位朋友们，大家好啊，今天此题我用的方法比较好理解，但时间复杂度比较高如果大家觉得可以的话，不妨给个免费的赞吧，谢谢了^ _ ^
1.题目要求如图所示:
在这里插入图片描述
2.做题步骤:
1.先计算总共多少个数:

    int count = 0;int number = 0;while(number <= n){count++;number++;}

2.然后再用malloc函数:

 int* number_t = (int*)malloc(sizeof(int) * count);

3.再采用双层while循环，再利按位与操作符和右移操作符来求每个数中1的个数:

int k = count;number = 0;count = 0;int f = 0;while(number <= n){int numbers = number;count = 0;while(numbers){if(numbers & 1 == 1){count++;}numbers =  numbers >> 1;}printf("%d ",count);number_t[f] = count;f++;number++;}

全部代码如图所示:

/*** Note: The returned array must be malloced, assume caller calls free().*/
int* countBits(int n, int* returnSize) {int count = 0;int number = 0;while(number <= n){count++;number++;}int* number_t = (int*)malloc(sizeof(int) * count);int k = count;number = 0;count = 0;int f = 0;while(number <= n){int numbers = number;count = 0;while(numbers){if(numbers & 1 == 1){count++;}numbers =  numbers >> 1;}printf("%d ",count);number_t[f] = count;f++;number++;}*returnSize = k;return number_t;   
}