• 桶排序(BucketSort)


    桶排序思想

    桶排序的思想就是把数组分到有限的桶中,然后再分别排序,最后将各个桶中的数据有序的合起来,具体如下:

    (1)设计好桶的大小和桶的个数

    (2)遍历数组,把每个元素放到对应的桶中

    (3)对每个桶分别排序(可以递归使用桶排序,也可以使用其它排序)

    (4)遍历桶中的数据,将数据有效的合起来

    复杂度

    桶排序利用函数映射关系,较少了几乎所有的比较操作。把大量数据分割成了基本有效的数据块,然后只需要对桶中的少量数据做先进的比较排序。

    对 n 个关键字进行桶排序的时间复杂度分为两部分:

    1、循环计算每个关键字的桶映射函数,这个时间复杂度是O(n).

    2、利用比较先进的比较排序算法,对每个桶中的数据进行排序,其时间复杂度是∑ O(Ni*logNi) 。其中Ni 为第i个桶的数据量。

    很显然,第(2)部分是桶排序性能好坏的决定因素。基于比较排序的最好平均时间复杂度只能达到O(N*logN),所以应该尽可能较少每个桶内数据的数量

    映射函数f(k)能够将N个数据平均的分配到M个桶中,这样每个桶就有[N/M]个数据量。

    所以尽量的增大桶的数量,极限情况下每个桶只能得到一个数据,这样就完全避开了桶内数据的“比较”排序操作。这就是一种空间换时间的策略。

    总结

    对于N个待排数据,M个桶,平均每个桶[N/M]个数据的桶排序平均时间复杂度为:

      O(N)+O(M*(N/M)*log(N/M))=O(N+N*(logN-logM))=O(N+N*logN-N*logM)

    当N=M时,即极限情况下每个桶只有一个数据时。桶排序的最好效率能够达到O(N)。

    桶排序的平均时间复杂度为线性的O(N+C),其中C=N*(logN-logM).

    空间复杂度O(N+M),如果输入数据非常庞大,而桶的数量也非常多,则空间代价无疑是昂贵的。

    此外,桶排序是稳定的

    代码实现

    简易版

     1 #include<stdio.h>
     2 #include<algorithm>
     3 #include<vector>
     4 using namespace std;
     5
     6 const int bucket_num = 10;    //桶的数量
     7 const int interval = 10;    //桶的容量
     8 const int maxn = 100 + 10;        //数字的最大值
     9 vector<int>buckets[bucket_num];    //每个桶
    10 
    11 void BucketSort(int *arr, int n)
    12 {
    13     for (int i = 0; i < n; i++)
    14         buckets[arr[i] / interval].push_back(arr[i]);
    15     for (int i = 0; i < bucket_num; i++)  if (buckets[i].size())
    16         sort(buckets[i].begin(), buckets[i].end());
    17     for (int i = 0; i < bucket_num; i++)  if (buckets[i].size())
    18         for (int j = 0; j < buckets[i].size(); j++)
    19             printf("%d ", buckets[i][j]);
    20     printf("
    ");
    21 }
    22 
    23 int main()
    24 {
    25     int n, arr[maxn];
    26     printf("数组大小:");
    27     scanf("%d", &n);
    28     printf("数组:");
    29     for (int i = 0; i < n; i++)  scanf("%d", &arr[i]);
    30     BucketSort(arr, n);
    31     
    32     return 0;
    33 }

    详细版

      1 #include<stdio.h>
      2 #include<stdlib.h>
      3 
      4 const int  NARRAY = 50 + 5;  /* array size */
      5 const int NBUCKET = 5; /* bucket size */
      6 const int INTERVAL = 10; /* bucket range */
      7 
      8 struct Node
      9 {
     10     int data;
     11     struct Node *next;
     12 };
     13 
     14 void BucketSort(int arr[],int n);
     15 struct Node *InsertionSort(struct Node *list);
     16 void print(int arr[],int n);
     17 void printBuckets(struct Node *list);
     18 int getBucketIndex(int value);
     19 
     20 void BucketSort(int arr[],int n)
     21 {
     22     struct Node **buckets;
     23 
     24     /* allocate memory for array of pointers to the buckets */
     25     buckets = (struct Node **)malloc(sizeof(struct Node*) * NBUCKET);
     26 
     27     /* initialize pointers to the buckets */
     28     for (int i = 0; i < NBUCKET; ++i) {
     29         buckets[i] = NULL;
     30     }
     31 
     32     /* put items into the buckets,like insert from the head */
     33     for (int i = 0; i < n; ++i) {
     34         struct Node *current;
     35         int pos = getBucketIndex(arr[i]);
     36         current = (struct Node *) malloc(sizeof(struct Node));
     37         current->data = arr[i];
     38         current->next = buckets[pos];
     39         buckets[pos] = current;
     40     }
     41 
     42     /* check what's in each bucket */
     43     for (int i = 0; i < NBUCKET; i++) {
     44         printf("Bucket[%d] : ", i);
     45         printBuckets(buckets[i]);
     46     }
     47 
     48     /* sorting bucket using Insertion Sort */
     49     for (int i = 0; i < NBUCKET; ++i) {
     50         buckets[i] = InsertionSort(buckets[i]);
     51     }
     52 
     53     /* check what's in each bucket */
     54     
     55     printf("-------------
    ");
     56     printf("Bucktets after sorted
    ");
     57 
     58     for (int i = 0; i < NBUCKET; i++) {
     59         printf("Bucket[%d] : ", i);
     60         printBuckets(buckets[i]);
     61     }
     62 
     63     /* put items back to original array */
     64     for (int j = 0, i = 0; i < NBUCKET; ++i) {
     65         struct Node *node;
     66         node = buckets[i];
     67         while (node) {
     68             arr[j++] = node->data;
     69             node = node->next;
     70         }
     71     }
     72 
     73     /* free memory */
     74     for (int i = 0; i < NBUCKET; ++i) {
     75         struct Node *node;
     76         node = buckets[i];
     77         while (node) {
     78             struct Node *tmp;
     79             tmp = node;
     80             node = node->next;
     81             free(tmp);
     82         }
     83     }
     84     free(buckets);
     85     return;
     86 }
     87 
     88 /* Insertion Sort */
     89 struct Node *InsertionSort(struct Node *list)  //list as head pionter
     90 {
     91     struct Node *k, *nodeList;
     92     /* need at least two items to sort */
     93     if (list == 0 || list->next == 0) {
     94         return list;
     95     }
     96 
     97     nodeList = list;
     98     k = list->next;
     99     nodeList->next = 0; /* 1st node is new list */
    100     while (k != 0) {
    101         struct Node *ptr;
    102         /* check if insert before first */
    103         if (nodeList->data > k->data) {
    104             struct Node *tmp;    //swap(k,nodeList)
    105             tmp = k;
    106             k = k->next;
    107             tmp->next = nodeList;
    108             nodeList = tmp;
    109             continue;
    110         }
    111 
    112         for (ptr = nodeList; ptr->next != 0; ptr = ptr->next) {
    113             if (ptr->next->data > k->data) break;
    114         }
    115 
    116         if (ptr->next != NULL) {  //if isn`t tail node,swap
    117             struct Node *tmp;
    118             tmp = k;
    119             k = k->next;
    120             tmp->next = ptr->next;
    121             ptr->next = tmp;
    122             continue;
    123         }
    124         else {                    //if isn tail node,insert at the tail
    125             ptr->next = k;
    126             k = k->next;
    127             ptr->next->next = 0;
    128             continue;
    129         }
    130     }
    131     return nodeList;
    132 }
    133 
    134 int getBucketIndex(int value)
    135 {
    136     return value / INTERVAL;
    137 }
    138 
    139 void print(int arr[],int n)
    140 {
    141     for (int i = 0; i < n; ++i) {
    142         printf("%3d",arr[i]);
    143     }
    144     printf("
    ");
    145 }
    146 
    147 void printBuckets(struct Node *list)
    148 {
    149     struct Node *cur = list;
    150     while (cur) {
    151         printf("%3d", cur->data);
    152         cur = cur->next;
    153     }
    154     printf("
    ");
    155 }
    156 
    157 int main()
    158 {
    159     int n, array[NARRAY];
    160     printf("数组大小:");
    161     scanf("%d", &n);        //less than NBUCKET*INTERVAL
    162     printf("数组:");
    163     for (int i = 0; i < n; i++)  scanf("%d", &array[i]);
    164 
    165     printf("Initial array
    ");
    166     print(array,n);
    167     printf("-------------
    ");
    168 
    169     BucketSort(array,n);
    170     printf("-------------
    ");
    171     printf("Sorted array
    ");
    172     print(array,n);
    173 
    174     return 0;
    175 }

     参考链接:

    https://baike.baidu.com/item/桶排序/4973777

    https://blog.csdn.net/u012194956/article/details/79887143                                                                                                                                                                                                                                                                                                                                                                                                                        

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  • 原文地址:https://www.cnblogs.com/lfri/p/9885218.html
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