利用了函数的映射关系,高效与否的关键就在于这个映射函数的确定。为了使桶排序更加高效,我们需要做到这两点:
- 在额外空间充足的情况下,尽量增大桶的数量
- 使用的映射函数能够将输入的 N 个数据均匀的分配到 K 个桶中
JavaScript
1 function bucketSort(arr, bucketSize) { 2 if (arr.length === 0) { 3 return arr; 4 } 5 6 var i; 7 var minValue = arr[0]; 8 var maxValue = arr[0]; 9 for (i = 1; i < arr.length; i++) { 10 if (arr[i] < minValue) { 11 minValue = arr[i]; // 输入数据的最小值 12 } else if (arr[i] > maxValue) { 13 maxValue = arr[i]; // 输入数据的最大值 14 } 15 } 16 17 //桶的初始化 18 var DEFAULT_BUCKET_SIZE = 5; // 设置桶的默认数量为5 19 bucketSize = bucketSize || DEFAULT_BUCKET_SIZE; 20 var bucketCount = Math.floor((maxValue - minValue) / bucketSize) + 1; 21 var buckets = new Array(bucketCount); 22 for (i = 0; i < buckets.length; i++) { 23 buckets[i] = []; 24 } 25 26 //利用映射函数将数据分配到各个桶中 27 for (i = 0; i < arr.length; i++) { 28 buckets[Math.floor((arr[i] - minValue) / bucketSize)].push(arr[i]); 29 } 30 31 arr.length = 0; 32 for (i = 0; i < buckets.length; i++) { 33 insertionSort(buckets[i]); // 对每个桶进行排序,这里使用了插入排序 34 for (var j = 0; j < buckets[i].length; j++) { 35 arr.push(buckets[i][j]); 36 } 37 } 38 39 return arr; 40 }
Python
1 def bucket_sort_simplify(arr, max_num): 2 buf = {i: [] for i in range(int(max_num)+1)} # 不能使用[[]]*(max+1),这样新建的空间中各个[]是共享内存的 3 arr_len = len(arr) 4 for i in range(arr_len): 5 num = arr[i] 6 buf[int(num)].append(num) # 将相应范围内的数据加入到[]中 7 arr = [] 8 for i in range(len(buf)): 9 if buf[i]: 10 arr.extend(sorted(buf[i])) # 这里还需要对一个范围内的数据进行排序,然后再进行输出 11 return arr 12 13 14 if __name__ == "__main__": 15 lis = [3.1, 4.2, 3.3, 3.5, 2.2, 2.7, 2.9, 2.1, 1.55, 4.456, 6.12, 5.2, 5.33, 6.0, 2.12] 16 print(bucket_sort_simplify(lis, max(lis)))
C++
1 #include<iterator> 2 #include<iostream> 3 #include<vector> 4 using namespace std; 5 const int BUCKET_NUM = 10; 6 7 struct ListNode{ 8 explicit ListNode(int i=0):mData(i),mNext(NULL){} 9 ListNode* mNext; 10 int mData; 11 }; 12 13 ListNode* insert(ListNode* head,int val){ 14 ListNode dummyNode; 15 ListNode *newNode = new ListNode(val); 16 ListNode *pre,*curr; 17 dummyNode.mNext = head; 18 pre = &dummyNode; 19 curr = head; 20 while(NULL!=curr && curr->mData<=val){ 21 pre = curr; 22 curr = curr->mNext; 23 } 24 newNode->mNext = curr; 25 pre->mNext = newNode; 26 return dummyNode.mNext; 27 } 28 29 30 ListNode* Merge(ListNode *head1,ListNode *head2){ 31 ListNode dummyNode; 32 ListNode *dummy = &dummyNode; 33 while(NULL!=head1 && NULL!=head2){ 34 if(head1->mData <= head2->mData){ 35 dummy->mNext = head1; 36 head1 = head1->mNext; 37 }else{ 38 dummy->mNext = head2; 39 head2 = head2->mNext; 40 } 41 dummy = dummy->mNext; 42 } 43 if(NULL!=head1) dummy->mNext = head1; 44 if(NULL!=head2) dummy->mNext = head2; 45 46 return dummyNode.mNext; 47 } 48 49 void BucketSort(int n,int arr[]){ 50 vector<ListNode*> buckets(BUCKET_NUM,(ListNode*)(0)); 51 for(int i=0;i<n;++i){ 52 int index = arr[i]/BUCKET_NUM; 53 ListNode *head = buckets.at(index); 54 buckets.at(index) = insert(head,arr[i]); 55 } 56 ListNode *head = buckets.at(0); 57 for(int i=1;i<BUCKET_NUM;++i){ 58 head = Merge(head,buckets.at(i)); 59 } 60 for(int i=0;i<n;++i){ 61 arr[i] = head->mData; 62 head = head->mNext; 63 } 64 }