归并排序（英语：Merge sort，或mergesort），是创建在归并操作上的一种有效的排序算法，效率为 � ( � log ⁡ � ) {\displaystyle 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。

目录 1 概述 2 归并操作 2.1 递归法（Top-down） 2.2 迭代法（Bottom-up） 3 实现示例 3.1 C语言 3.2 C++ 3.3 C# 3.4 Ruby 3.5 Java 3.6 PHP 3.7 Python3 3.8 Erlang 3.9 Javascript 3.10 Go 4 算法复杂度 5 参考文献 6 外部链接概述采用分治法:

分割：递归地把当前序列平均分割成两半。集成：在保持元素顺序的同时将上一步得到的子序列集成到一起（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 � � � � ( � / 2 ) {\displaystyle 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 � � � � ( � / 4 ) {\displaystyle 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 实现示例 C语言迭代版：

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); } 递归版：

// 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 // 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; } 递归版：

void Merge(vector &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 LeftSubArray(Array.begin() + front, Array.begin() + mid + 1); vector RightSubArray(Array.begin() + mid + 1, Array.begin() + end + 1); int idxLeft = 0, idxRight = 0; LeftSubArray.insert(LeftSubArray.end(), numeric_limits::max()); RightSubArray.insert(RightSubArray.end(), numeric_limits::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 &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); } [1]

C# public static List sort(List lst) { if (lst.Count <= 1) return lst; int mid = lst.Count / 2; List left = new List(); // 定义左侧List List right = new List(); // 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); } ///

/// 合併兩個已經排好序的List ///

/// 左側List /// 右側List /// static List merge(List left, List right) { List temp = new List(); 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 递归版：

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 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 ( � log ⁡ � ) / 2 {\displaystyle (n\log n)/2}和 � log ⁡ � − � + 1 {\displaystyle n\log n-n+1}。赋值操作的次数是 ( 2 � log ⁡ � ) {\displaystyle (2n\log n)}。归并算法的空间复杂度为： Θ ( � ) {\displaystyle \Theta (n)}

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