LeetCode 338.比特位计数

付宇轩2024-06-20 22:32

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:

2.做题步骤:

1.先计算总共多少个数:

c 复制代码

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

2.然后再用malloc函数:

c 复制代码

 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:

c 复制代码

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:

c 复制代码

/**
 * 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;   
}

好了，这就是我全部的代码，大家如果觉得好的话，请给个免费的赞吧，谢谢了^ _ ^