数据结构第六次作业第三题（桶排序）-优快云博客

本文链接：https://blog.youkuaiyun.com/MysteriousL/article/details/109685725

该代码实现了一个数组的排序算法，结合了桶分配策略。首先读取一串数字，然后根据输入需求进行排序。排序过程中使用了二分查找优化，并通过动态调整桶大小来适应不同规模的数据。最后，如果满足特定条件则打印排序结果。

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#include <stdio.h>

int min = -99999;

void print(int a[], int n)
{
    for (int i = 0; i < n; i++)
        printf("%d,", a[i]);
}

void sort(int a[], int par, int size)
{
    int temp = a[par];
    int child = 2 * par + 1;
    while (child < size)
    {
        if (child + 1 < size && a[child + 1] > a[child])
            child++;
        if (a[par] > a[child])
            break;
        else
        {
            a[par] = a[child];
            par = child;
            child = 2 * par + 1;
        }
        a[par] = temp;
    }
}

void swap(int a[], int i, int j)
{
    int temp = a[i];
    a[i] = a[j];
    a[j] = temp;
}

int main()
{
    char b[50];
    gets(b);
    int a[21];
    int bucket[10][21];
    for (int i = 0; i < 10; i++)
        for (int j = 0; j < 21; j++)
            bucket[i][j] = min;
    int sum = 0, run = 0, i = 0, cnt = 0;
    while (b[i])
    {
        if (b[i] != ',')
        {
            sum = sum * 10 + b[i] - '0';
        }
        else
        {
            a[run++] = sum;
            sum = 0;
        }
        i++;
    }
    int need;
    scanf("%d", &need);
    if (cnt == need)
        print(a, run);
    int pr = 10;
    int flag = 0;
    int row = 0, column = 0;
    while (1)
    {
        flag = 0;
        for (int i = 0; i < run; i++)
        {
            int k = (a[i] % pr) / (pr / 10);
            // printf("k=%d\n", k);
            bucket[k][column++] = a[i];
            if (a[i] / pr != 0 && a[i] != 0)
            {
                flag = 1;
            }
        }
        int run = 0;
        for (int i = 0; i < 10; i++)
        {
            for (int j = 0; j < 21; j++)
            {
                if (bucket[i][j] != min)
                {
                    a[run++] = bucket[i][j];
                }
            }
        }
        cnt++;
        if (cnt == need)
            print(a, run);
        if (flag == 0)
            break;
        for (int i = 0; i < 10; i++)
            for (int j = 0; j < 21; j++)
                bucket[i][j] = min;
        pr *= 10;
    }

    cnt++;
    if (cnt == need)
        print(a, run);

    return 0;
}