桶排序

$O(n+k)$，$k$为桶数

$$O(n\cdot k)$$

$$O(n)$$

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桶排序（Bucket sort）或所谓的箱排序，是一个排序算法，工作的原理是将阵列分到有限数量的桶里。每个桶再个别排序（有可能再使用别的排序算法或是以递回方式继续使用桶排序进行排序）。桶排序是鸽巢排序的一般化(generalization)。当要被排序的阵列内的数值是均匀分配的时候，桶排序使用线性时间（$\Theta(n)$）。

桶排序以下列程序进行：

- 设置一个定量的阵列当作空桶子。
- 寻访序列，并且把项目一个一个放到对应的桶子去。
- 对每个不是空的桶子进行排序。
- 从不是空的桶子里把项目再放回原来的序列中。

## 伪代码

```
'''function''' bucket-sort(array, n) '''is'''
  buckets ← new array of n empty lists
  '''for''' i = 0 '''to''' (length(array)-1) '''do'''
    insert ''array[i]'' into buckets[msbits(array[i], k)]
  '''for''' i = 0 '''to''' n - 1 '''do'''
    next-sort(buckets[i])
  '''return''' the concatenation of buckets[0], ..., buckets[n-1]
```

## C++实现算法

假设数据分布在0，100)之间，每个桶内部用链表表示，在数据入桶的同时插入排序。然后把各个桶中的数据合并。

```cpp
#include<iterator>
#include<iostream>
#include<vector>
using namespace std;
const int BUCKET_NUM = 10;

struct ListNode{
	explicit ListNode(int i=0):mData(i),mNext(NULL){}
	ListNode* mNext;
	int mData;
};

ListNode* insert(ListNode* head,int val){
	ListNode dummyNode;
	ListNode *newNode = new ListNode(val);
	ListNode *pre,*curr;
	dummyNode.mNext = head;
	pre = &dummyNode;
	curr = head;
	while(NULL!=curr && curr->mData<=val){
		pre = curr;
		curr = curr->mNext;
	}
	newNode->mNext = curr;
	pre->mNext = newNode;
	return dummyNode.mNext;
}

ListNode* Merge(ListNode *head1,ListNode *head2){
	ListNode dummyNode;
	ListNode *dummy = &dummyNode;
	while(NULL!=head1 && NULL!=head2){
		if(head1->mData <= head2->mData){
			dummy->mNext = head1;
			head1 = head1->mNext;
		}else{
			dummy->mNext = head2;
			head2 = head2->mNext;
		}
		dummy = dummy->mNext;
	}
	if(NULL!=head1) dummy->mNext = head1;
	if(NULL!=head2) dummy->mNext = head2;

	return dummyNode.mNext;
}

void BucketSort(int n,int arr[]){
	vector<ListNode*> buckets(BUCKET_NUM,(ListNode*)(0));
	for(int i=0;i<n;++i){
		int index = arr[i]/BUCKET_NUM;
		ListNode *head = buckets.at(index);
		buckets.at(index) = insert(head,arr[i]);
	}
	ListNode *head = buckets.at(0);
	for(int i=1;i<BUCKET_NUM;++i){
		head = Merge(head,buckets.at(i));
	}
	for(int i=0;i<n;++i){
		arr[i] = head->mData;
		head = head->mNext;
	}
}
```

## Java实现算法

```java
	private int indexFor(int a, int min, int step) {
		return (a - min) / step;
	}

	public void bucketSort(int[] arr) {

		int max = arr[0], min = arr[0];
		for (int a : arr) {
			if (max < a)
				max = a;
			if (min > a)
				min = a;
		}
		// 該值也可根據實際情況選擇
		int bucketNum = max / 10 - min / 10 + 1;
		List buckList = new ArrayList<List<Integer>>();
		// create bucket
		for (int i = 1; i <= bucketNum; i++) {
			buckList.add(new ArrayList<Integer>());
		}
		// push into the bucket
		for (int i = 0; i < arr.length; i++) {
			int index = indexFor(arr[i], min, 10);
			((ArrayList<Integer>) buckList.get(index)).add(arr[i]);
		}
		ArrayList<Integer> bucket = null;
		int index = 0;
		for (int i = 0; i < bucketNum; i++) {
			bucket = (ArrayList<Integer>) buckList.get(i);
			insertSort(bucket);
			for (int k : bucket) {
				arr[index++] = k;
			}
		}

	}

	// 把桶內元素插入排序
	private void insertSort(List<Integer> bucket) {
		for (int i = 1; i < bucket.size(); i++) {
			int temp = bucket.get(i);
			int j = i - 1;
			for (; j >= 0 && bucket.get(j) > temp; j--) {
				bucket.set(j + 1, bucket.get(j));
			}
			bucket.set(j + 1, temp);
		}
	}
```

## JavaScript实现算法

```ecmascript
Array.prototype.bucketSort = function(num) {
  function swap(arr, i, j) {
    const temp = arr[i]
    arr[i] = arr[j]
    arr[j] = temp
  }
  const max = Math.max(...this)
  const min = Math.min(...this)
  const buckets = []
  const bucketsSize = Math.floor((max - min) / num) + 1
  for(let i = 0; i < this.length; i++) {
    const index = ~~(this[i] / bucketsSize)
    !buckets[index] && (buckets[index] = [])
    buckets[index].push(this[i])
    let l = buckets[index].length
    while(l > 0) {
      buckets[index][l] < buckets[index][l - 1] && swap(buckets[index], l, l - 1)
      l--
    }
  }
  let wrapBuckets = []
  for(let i = 0; i < buckets.length; i++) {
    buckets[i] && (wrapBuckets = wrapBuckets.concat(buckets[i]))
  }
  return wrapBuckets
}
const arr = [11, 9, 6, 8, 1, 3, 5, 1, 1, 0, 100]
console.log(arr.bucketSort(10))
```

## JavaScript实现递归算法

```ecmascript
Array.prototype.bucketSort = function()
{
  var start = 0;
  var size = this.length;
  var min = this[0];
  var max = this[0]; 
  for (var i = 1; i < size; i++){
    if (this[i] < min){min = this[i];}
    else{ if(this[i] > max){max = this[i];} }
  }
  if (min != max){
    var bucket = new Array(size);
    for (var i = 0; i < size; i++){bucket[i] = new Array();}
    var interpolation = 0;
    for (var i = 0; i < size; i++){
      interpolation = Math.floor(((this[i] - min) / (max - min)) * (size - 1));
      bucket[interpolation].push(this[i]);
    }
    for (var i = 0; i < size; i++){
      if (bucket[i].length > 1){bucket[i].bucketSort();}//遞歸
      for(var j = 0; j < bucket[i].length; j++){this[start++] = bucket[i][j];}
    }
  }
};  
```

## Python 3.10 实现算法

```python
def is_sort(arr: list) -> bool:
    for i in range(len(arr) - 1):
        if arr[i] > arr[i + 1]:
            return False
    return True

def bucket_sort(arr: list, is_sub_bucket: bool = False):
    if is_sort(arr):
        return

    bucket_num: int = max(arr) // 10 + 1 if not is_sub_bucket else 10
    buckets: list[list] = [[] for _ in range(bucket_num)]

    for a in arr:
        i: int = a // 10 if not is_sub_bucket else a % 10
        buckets[i].append(a)
    
    arr.clear()
    for bucket in buckets:
        bucket_sort(bucket, is_sub_bucket=True)
        arr += bucket

if __name__ == '__main__':
    arr = [29, 25, 3, 49, 9, 37, 21, 43]
    bucket_sort(arr)
    print(arr)
```

## 外部链接

- Bucket Sort in C++
- Bucket Sort & PHP Sample

## 参考

- Thomas H. Cormen，Charles E. Leiserson，Ronald L. Rivest，and Clifford Stein。Introduction to Algorithms, Second Edition. MIT Press and McGraw-Hill, 2001. ISBN 0-262-03293-7. Section 8.4: Bucket sort, pp.174–177.
- Paul E. Black "Postman's Sort" from Dictionary of Algorithms and Data Structures at NIST。
- Robert Ramey The Postman's Sort C Users Journal Aug. 1992

## 基本资料

- 车辆类别：排序算法
- data：数组
- 时间：$O(n^2)$
- average-time：$O(n+k)$，$k$为桶数
- 容积、占地面积：$O(n\cdot k)$
- optimal：$O(n)$