归并排序

归并排序
归并排序
	使用合併排序為一列數字進行排序的過程
概况
類別	排序算法
資料結構	数组
复杂度
平均時間複雜度
最坏时间复杂度
最优时间复杂度
空間複雜度
最佳解	有时是
相关变量的定义

归并排序（英語：Merge sort，或mergesort），是建立在归并操作上的一种有效的排序算法，效率為 $O(n\log n)$ （大O符号）。1945年由约翰·冯·诺伊曼首次提出。该算法是采用分治法（Divide and Conquer）的一个非常典型的应用，且各层分治递归可以同时进行。

概述

采用分治法:

分割：递归地把当前序列平均分割成两半。
整合：在保持元素顺序的同时将上一步得到的子序列整合到一起（归并）。

归并操作

归并操作（merge），也叫归并算法，指的是将两个已经排序的序列合并成一个序列的操作。归并排序算法依赖归并操作。

递归法（Top-down）

申请空间，使其大小为两个已经排序序列之和，该空间用来存放合并后的序列
设定两个指针，最初位置分别为两个已经排序序列的起始位置
比较两个指针所指向的元素，选择相对小的元素放入到合并空间，并移动指针到下一位置
重复步骤3直到某一指针到达序列尾
将另一序列剩下的所有元素直接复制到合并序列尾

迭代法（Bottom-up）

原理如下（假设序列共有 $n$ 个元素）：

将序列每相邻两个数字进行归并操作，形成 $ceil(n/2)$ 个序列，排序后每个序列包含两/一个元素
若此时序列数不是1个则将上述序列再次归并，形成 $ceil(n/4)$ 个序列，每个序列包含四/三个元素
重复步骤2，直到所有元素排序完毕，即序列数为1

實作範例

C語言

迭代版：

int min(int x, int y) {
    return x < y ? x : y;
}
void merge_sort(int arr[], int len) {
    int *a = arr;
    int *b = (int *) malloc(len * sizeof(int));
    int seg, start;
    for (seg = 1; seg < len; seg += seg) {
        for (start = 0; start < len; start += seg * 2) {
            int low = start, mid = min(start + seg, len), high = min(start + seg * 2, len);
            int k = low;
            int start1 = low, end1 = mid;
            int start2 = mid, end2 = high;
            while (start1 < end1 && start2 < end2)
                b[k++] = a[start1] < a[start2] ? a[start1++] : a[start2++];
            while (start1 < end1)
                b[k++] = a[start1++];
            while (start2 < end2)
                b[k++] = a[start2++];
        }
        int *temp = a;
        a = b;
        b = temp;
    }
    if (a != arr) {
        int i;
        for (i = 0; i < len; i++)
            b[i] = a[i];
        b = a;
    }
    free(b);
}

遞歸版：

// 分治-治
void mergeSort_conquer(int* array, int left, int mid, int right, int* temp) {
    // [left, mid]和[mid+1, right]两个有序数组
    int i = left;
    int j = mid + 1;
    int index = 0;
    while (i <= mid && j <= right) {
        if (array[i] < array[j]) {
            temp[index++] = array[i++];
        } else {
            temp[index++] = array[j++];
        }
    }
    // 剩余元素直接放入temp
    while (i <= mid) {
        temp[index++] = array[i++];
    }
    while (j <= right) {
        temp[index++] = array[j++];
    }
    // 放回原数组
    index = 0;
    while (left <= right) {
        array[left++] = temp[index++];
    }
}

// 分治-分
void mergeSort_divide(int* array, int left, int right, int* temp) {
    if (left < right) {
        int mid = left + (right - left) / 2;
        // 左边归并排序
        mergeSort_divide(array, left, mid, temp);
        // 右边归并排序
        mergeSort_divide(array, mid + 1, right, temp);
        // 合并两个有序序列
        mergeSort_conquer(array, left, mid, right, temp);
    }
}

void mergeSort(int* array, int size) {
    int* temp = (int*)malloc(sizeof(int) * size);
    mergeSort_divide(array, 0, size - 1, temp);
}

C++

迭代版：

template<typename T> // 整數或浮點數皆可使用,若要使用物件(class)時必須設定"小於"(<)的運算子功能
void merge_sort(T arr[], int len) {
    T *a = arr;
    T *b = new T[len];
    for (int seg = 1; seg < len; seg += seg) {
        for (int start = 0; start < len; start += seg + seg) {
            int low = start, mid = min(start + seg, len), high = min(start + seg + seg, len);
            int k = low;
            int start1 = low, end1 = mid;
            int start2 = mid, end2 = high;
            while (start1 < end1 && start2 < end2)
                b[k++] = a[start1] < a[start2] ? a[start1++] : a[start2++];
            while (start1 < end1)
                b[k++] = a[start1++];
            while (start2 < end2)
                b[k++] = a[start2++];
        }
        T *temp = a;
        a = b;
        b = temp;
    }
    if (a != arr) {
        for (int i = 0; i < len; i++)
            b[i] = a[i];
        b = a;
    }
    delete[] b;
}

遞歸版：

void Merge(vector<int> &Array, int front, int mid, int end) {
    // preconditions:
    // Array[front...mid] is sorted
    // Array[mid+1 ... end] is sorted
    // Copy Array[front ... mid] to LeftSubArray
    // Copy Array[mid+1 ... end] to RightSubArray
    vector<int> LeftSubArray(Array.begin() + front, Array.begin() + mid + 1);
    vector<int> RightSubArray(Array.begin() + mid + 1, Array.begin() + end + 1);
    int idxLeft = 0, idxRight = 0;
    LeftSubArray.insert(LeftSubArray.end(), numeric_limits<int>::max());
    RightSubArray.insert(RightSubArray.end(), numeric_limits<int>::max());
    // Pick min of LeftSubArray[idxLeft] and RightSubArray[idxRight], and put into Array[i]
    for (int i = front; i <= end; i++) {
        if (LeftSubArray[idxLeft] < RightSubArray[idxRight]) {
            Array[i] = LeftSubArray[idxLeft];
            idxLeft++;
        } else {
            Array[i] = RightSubArray[idxRight];
            idxRight++;
        }
    }
}

void MergeSort(vector<int> &Array, int front, int end) {
    if (front >= end)
        return;
    int mid = front + (end - front) / 2;
    MergeSort(Array, front, mid);
    MergeSort(Array, mid + 1, end);
    Merge(Array, front, mid, end);
}

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C#

public static List<int> sort(List<int> lst) {
    if (lst.Count <= 1)
        return lst;
    int mid = lst.Count / 2;
    List<int> left = new List<int>();  // 定义左侧List
    List<int> right = new List<int>(); // 定义右侧List
    // 以下兩個循環把 lst 分為左右兩個 List
    for (int i = 0; i < mid; i++)
        left.Add(lst[i]);
    for (int j = mid; j < lst.Count; j++)
        right.Add(lst[j]);
    left = sort(left);
    right = sort(right);
    return merge(left, right);
}
/// <summary>
/// 合併兩個已經排好序的List
/// </summary>
/// <param name="left">左側List</param>
/// <param name="right">右側List</param>
/// <returns></returns>
static List<int> merge(List<int> left, List<int> right) {
    List<int> temp = new List<int>();
    while (left.Count > 0 && right.Count > 0) {
        if (left[0] <= right[0]) {
            temp.Add(left[0]);
            left.RemoveAt(0);
        } else {
            temp.Add(right[0]);
            right.RemoveAt(0);
        }
    }
    if (left.Count > 0) {
        for (int i = 0; i < left.Count; i++)
            temp.Add(left[i]);
    }
    if (right.Count > 0) {
        for (int i = 0; i < right.Count; i++)
            temp.Add(right[i]);
    }
    return temp;
}

Ruby

def merge list
  return list if list.size < 2

  pivot = list.size / 2

  # Merge
  lambda { |left, right|
    final = []
    until left.empty? or right.empty?
      final << if left.first < right.first; left.shift else right.shift end
    end
    final + left + right
  }.call merge(list[0...pivot]), merge(list[pivot..-1])
end

Java

遞歸版：

static void merge_sort_recursive(int[] arr, int[] result, int start, int end) {
	if (start >= end)
		return;
	int len = end - start, mid = (len >> 1) + start;
	int start1 = start, end1 = mid;
	int start2 = mid + 1, end2 = end;
	merge_sort_recursive(arr, result, start1, end1);
	merge_sort_recursive(arr, result, start2, end2);
	int k = start;
	while (start1 <= end1 && start2 <= end2)
		result[k++] = arr[start1] < arr[start2] ? arr[start1++] : arr[start2++];
	while (start1 <= end1)
		result[k++] = arr[start1++];
	while (start2 <= end2)
		result[k++] = arr[start2++];
	for (k = start; k <= end; k++)
		arr[k] = result[k];
}
public static void merge_sort(int[] arr) {
	int len = arr.length;
	int[] result = new int[len];
	merge_sort_recursive(arr, result, 0, len - 1);
}

迭代版：

public static void merge_sort(int[] arr) {
	int[] orderedArr = new int[arr.length];
	for (int i = 2; i < arr.length * 2; i *= 2) {
		for (int j = 0; j < (arr.length + i - 1) / i; j++) {
			int left = i * j;
			int mid = left + i / 2 >= arr.length ? (arr.length - 1) : (left + i / 2);
			int right = i * (j + 1) - 1 >= arr.length ? (arr.length - 1) : (i * (j + 1) - 1);
			int start = left, l = left, m = mid;
			while (l < mid && m <= right) {
				if (arr[l] < arr[m]) {
					orderedArr[start++] = arr[l++];
				} else {
					orderedArr[start++] = arr[m++];
				}
			}
			while (l < mid)
				orderedArr[start++] = arr[l++];
			while (m <= right)
				orderedArr[start++] = arr[m++];
			System.arraycopy(orderedArr, left, arr, left, right - left + 1);
		}
	}
}

PHP

function merge_sort($arr) {
	$len = count($arr);
	if ($len <= 1)
		return $arr;
	$half = ($len>>1) + ($len & 1);
	$arr2d = array_chunk($arr, $half);
	$left = merge_sort($arr2d[0]);
	$right = merge_sort($arr2d[1]);
	while (count($left) && count($right))
		if ($left[0] < $right[0])
			$reg[] = array_shift($left);
		else
			$reg[] = array_shift($right);
	return array_merge($reg, $left, $right);
}

$arr = array(21, 34, 3, 32, 82, 55, 89, 50, 37, 5, 64, 35, 9, 70);
$arr = merge_sort($arr);
for ($i = 0; $i < count($arr); $i++) {
	echo $arr[$i] . ' ';
}

Python3

def mergeSort(nums):
    if len(nums) < 2:
        return nums
    mid = len(nums) // 2
    left = mergeSort(nums[:mid])
    right = mergeSort(nums[mid:])

    i = j = 0
    result = []
    while i < len(left) and j < len(right):
        if left[i] < right[j]: 
            result.append(left[i])
            i += 1
        else: 
            result.append(right[j])
            j += 1

    while i < len(left): 
        result.append(left[i]) 
        i += 1

    while j < len(right): 
        result.append(right[j]) 
        j += 1

    return result


if __name__ == "__main__":
    nums = [1, 4, 2, 3.6, -1, 0, 25, -34, 8, 9, 1, 0]
    print("original:", nums)
    print("Sorted:", mergeSort(nums))

Erlang

%% @doc 归并排序
g_sort([]) ->
    [];
g_sort([T]) ->
    [T];
g_sort(L) ->
    g_sort(L, length(L)).

g_sort([_, _ | _] = L, Length) ->
    SplitNum = trunc(Length / 2),
    {L1, L2} = lists:split(SplitNum, L),
    g_merge(g_sort(L1, SplitNum), g_sort(L2, Length - SplitNum));
g_sort(L, _Length) ->
    L.

%% 已经排好序的两个list合并
g_merge([], L2) ->
    L2;
g_merge(L1, []) ->
    L1;
g_merge([T1 | Rest1] = L1, [T2 | Rest2] = L2) ->
    if
        T1 =< T2 -> [T1 | g_merge(Rest1, L2)];
        true -> [T2 | g_merge(L1, Rest2)]
    end.

Javascript

递归法

function merge(left, right){
  var result = [];
  while(left.length > 0 && right.length > 0){
    if(left[0] < right[0]){
      result.push(left.shift());
    }else{
      result.push(right.shift());
    }
  }
  return result.concat(left, right);
}

function mergeSort(arr){
  if(arr.length <=1) return arr;
  var middle = Math.floor(arr.length / 2);
  var left = arr.slice(0, middle);
  var right = arr.slice(middle);
  return merge(mergeSort(left), mergeSort(right));
}

迭代法

Go

package main

import (
	"fmt"
	"sort"
)

func MergeSort(list []int) []int {
	var length = len(list)
	if length < 2 {
		return list
	}
	var mid = length / 2
	return merge(MergeSort(list[:mid]), MergeSort(list[mid:]))
}

func merge(x, y []int) []int {
	var r []int = make([]int, len(x)+len(y))
	for i, j := 0, 0; ; {
		if i < len(x) && (j == len(y) || x[i] < y[j]) {
			r[i+j] = x[i]
			i++
		} else if j < len(y) {
			r[i+j] = y[j]
			j++
		} else {
			break
		}
	}
	return r
}

func main() {
	var list []int = []int{56, 48, 58, 94, 87, 4, 5, 61, 5, 8, 74, 9, 84, 15, 94, 9, 4, 31, 41, 68, 7, 4, 6, 94, 16, 9, 8, 4}
	fmt.Println(MergeSort(list))
	fmt.Println(list)

	sort.Ints(list)
	fmt.Println(list)
}

递归版

package main

import (
	"fmt"
)

func merge(data []int) []int {
	sum := len(data)
	if sum <= 1 {
		return data
	}
	left := data[0 : sum/2]
	lSize := len(left)
	if lSize >= 2 {
		left = merge(left)
	}
	right := data[sum/2:]
	rSize := len(right)
	if rSize >= 2 {
		right = merge(right)
	}
	j := 0
	t := 0
	arr := make([]int, sum)
	fmt.Println(left, right, data)
	for i := 0; i < sum; i++ {
		if j < lSize && t < rSize {
			if left[j] <= right[t] {
				arr[i] = left[j]
				j++
			} else {
				arr[i] = right[t]
				t++
			}	
		}  else if j >= lSize{
			arr[i] = right[t]
			t++
		}  else if t >= rSize{
			arr[i] = left[j]
			j++
		}
	}
	return arr
}

func main() {
	var aa = []int{1000, 2, 31, 34, 5, 9, 7, 4, 6, 89, 90, 99, 99, 99, 99, 99}
	
	var bb = merge(aa)
	fmt.Println(bb)
}

算法复杂度

比较操作的次数介于 $(n\log n)/2$ 和 $n\log n-n+1$ 。

赋值操作的次数是 $(2n\log n)$ 。归并算法的空间复杂度为： $\Theta (n)$

参考文献

^ Cormen, Thomas H. （英语：Thomas H. Cormen）; Leiserson, Charles E. （英语：Charles E. Leiserson）; Rivest, Ronald L.; Stein, Clifford. Section 2.3: Designing algorithms. Introduction to Algorithms 3rd. MIT Press and McGraw-Hill. 2009: 29–34 [1990]. ISBN 0-262-03384-4. .

外部連結

Dictionary of Algorithms and Data Structures: Merge sort （页面存档备份，存于互联网档案馆）
Merge sort Algorithm with example program （页面存档备份，存于互联网档案馆）
Merge Sort Algorithm Simulation
Mergesort For Linked Lists （页面存档备份，存于互联网档案馆）
Mergesort in Java, Python, Perl, C, Ruby

[1] Cormen, Thomas H. （英语：Thomas H. Cormen）; Leiserson, Charles E. （英语：Charles E. Leiserson）; Rivest, Ronald L.; Stein, Clifford. Section 2.3: Designing algorithms. Introduction to Algorithms 3rd. MIT Press and McGraw-Hill. 2009: 29–34 [1990]. ISBN 0-262-03384-4. .

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