桶排序

桶排序是計數(shù)排序的升級版。它利用了函數(shù)的映射關系，高效與否的關鍵就在于這個映射函數(shù)的確定懦傍。為了使桶排序更加高效丙笋，我們需要做到這兩點：

• 在額外空間充足的情況下谢澈，盡量增大桶的數(shù)量
• 使用的映射函數(shù)能夠將輸入的 N 個數(shù)據(jù)均勻的分配到 K 個桶中

同時，對于桶中元素的排序御板，選擇何種比較排序算法對于性能的影響至關重要锥忿。

1.什么時候最快

當輸入的數(shù)據(jù)可以均勻的分配到每一個桶中。

2.什么時候最慢

當輸入的數(shù)據(jù)被分配到了同一個桶中稳吮。

3.示意圖

元素分布在桶中：

bucket_sort_1.svg_.png

然后缎谷，元素在每個桶中排序：

bucket_sort_2.svg_.png

4.代碼

#include <stdio.h>
#include <stdlib.h>

#define BUCKET_NUM 10

struct ListNode {
    struct ListNode* next;
    int data;
};

struct ListNode*
insert(struct ListNode *head, int value)
{
    struct ListNode dummyNode;
    struct ListNode *newNode = (struct ListNode *)malloc(sizeof(struct ListNode));
    struct ListNode *pre, *curr;

    newNode->data = value;
    dummyNode.next = head;
    pre = &dummyNode;
    curr = head;

    while (curr != NULL && curr->data <= value) {
        pre = curr;
        curr = curr->next;
    }
    newNode->next = curr;
    pre->next = newNode;
    return dummyNode.next;
}

struct ListNode*
merge(struct ListNode *head1, struct ListNode *head2)
{
    struct ListNode dummyNode;
    struct ListNode *dummy = &dummyNode;

    while (head1 != NULL && head2 != NULL) {
        if (head1->data <= head2->data) {
            dummy->next = head1;
            head1 = head1->next;
        } else {
            dummy->next = head2;
            head2 = head2->next;
        }
        dummy = dummy->next;
    }

    if (head1 != NULL) dummy->next = head1;
    if (head2 != NULL) dummy->next = head2;
    return dummyNode.next;
}

void
bucket_sort(int n, int arr[])
{
    struct ListNode* buckets[BUCKET_NUM] = {0};

    int i;
    for (i = 0; i < n; ++i) {
        int index = arr[i] / BUCKET_NUM;
        struct ListNode *head = buckets[index];
        buckets[index] = insert(head, arr[i]);
    }

    struct ListNode *head = buckets[0];
    for (i = 1; i < BUCKET_NUM; ++i) {
        head = merge(head, buckets[i]);
    }

    for (i = 0; i < n; ++i) {
        arr[i] = head->data;
        head = head->next;
    }
}

int
main()
{
    int arr[] = {22, 34, 3, 32, 82, 55, 89, 50, 37, 5, 64, 35, 9, 70};
    int len = (int) sizeof(arr) / sizeof(*arr);

    bucket_sort(len, arr);

    int i;
    for (i = 0; i < len; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");
    return 0;
}

部分內容轉載于菜鳥教程

2021.01.29 16:47 深圳