归并排序（MERGE-SORT）是利用归并的思想实现的排序方法，该算法采用经典的分治（divide-and-conquer）策略（分治法将问题分(divide)成一些小的问题然后递归求解，而**治(conquer)**的阶段则将分的阶段得到的各答案”修补”在一起，即分而治之)，下面为大家详细讲解一下归并排序。
作为一种典型的分而治之思想的算法应用，归并排序的实现由两种方法：
自上而下的递归（所有递归的方法都可以用迭代重写，所以就有了第 2 种方法）； 自下而上的迭代；
在《数据结构与算法 JavaScript 描述》中，作者给出了自下而上的迭代方法。但是对于递归法，作者却认为：
However, it is not possible to do so in JavaScript, as the recursion goes too deep for the language to handle.
然而，在 JavaScript 中这种方式不太可行，因为这个算法的递归深度对它来讲太深了。
说实话，我不太理解这句话。意思是 JavaScript 编译器内存太小，递归太深容易造成内存溢出吗？还望有大神能够指教。
和选择排序一样，归并排序的性能不受输入数据的影响，但表现比选择排序好的多，因为始终都是 O(nlogn) 的时间复杂度。代价是需要额外的内存空间。
算法步骤
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 直到某一指针达到序列尾； 将另一序列剩下的所有元素直接复制到合并序列尾。
动图演示
代码实现
JavaScript
实例
function mergeSort(arr) { // 采用自上而下的递归方法
var len = arr.length;
if(len return arr;
}
var middle = Math.floor(len / 2),
left = arr.slice(0, middle),
right = arr.slice(middle);
return merge(mergeSort(left), mergeSort(right));
}
function merge(left, right)
{
var result = [];
while (left.length && right.length) {
if (left[0] else {
result.push(right.shift());
}
}
while (left.length)
result.push(left.shift());
while (right.length)
result.push(right.shift());
return result;
}
Python
实例
def mergeSort(arr):
import math
if(len(arr)return arr
middle = math.floor(len(arr)/2)
left, right = arr[0:middle], arr[middle:]
return merge(mergeSort(left), mergeSort(right))
def merge(left,right):
result = []
while left and right:
if left[0] else:
result.append(right.pop(0));
while left:
result.append(left.pop(0))
while right:
result.append(right.pop(0));
return result
Go
实例
func mergeSort(arr []int) []int {
length := len(arr)
if length return arr
}
middle := length / 2
left := arr[0:middle]
right := arr[middle:]
return merge(mergeSort(left), mergeSort(right))
}
func merge(left []int, right []int) []int {
var result []int
for len(left) != 0 && len(right) != 0 {
if left[0] else {
result = append(result, right[0])
right = right[1:]
}
}
for len(left) != 0 {
result = append(result, left[0])
left = left[1:]
}
for len(right) != 0 {
result = append(result, right[0])
right = right[1:]
}
return result
}
Java
实例
public class MergeSort implements IArraySort {
@Override
public int[] sort(int[] sourceArray) throws Exception {
// 对 arr 进行拷贝，不改变参数内容
int[] arr = Arrays.copyOf(sourceArray, sourceArray.length);
if (arr.length return arr;
}
int middle = (int) Math.floor(arr.length / 2);
int[] left = Arrays.copyOfRange(arr, 0, middle);
int[] right = Arrays.copyOfRange(arr, middle, arr.length);
return merge(sort(left), sort(right));
}
protected int[] merge(int[] left, int[] right) {
int[] result = new int[left.length + right.length];
int i = 0;
while (left.length > 0 && right.length > 0) {
if (left[0] else {
result[i++] = right[0];
right = Arrays.copyOfRange(right, 1, right.length);
}
}
while (left.length > 0) {
result[i++] = left[0];
left = Arrays.copyOfRange(left, 1, left.length);
}
while (right.length > 0) {
result[i++] = right[0];
right = Arrays.copyOfRange(right, 1, right.length);
}
return result;
}
}
PHP
实例
function mergeSort($arr)
{
$len = count($arr);
if ($len return $arr;
}
$middle = floor($len / 2);
$left = array_slice($arr, 0, $middle);
$right = array_slice($arr, $middle);
return merge(mergeSort($left), mergeSort($right));
}
function merge($left, $right)
{
$result = [];
while (count($left) > 0 && count($right) > 0) {
if ($left[0] $right[0]) {
$result[] = array_shift($left);
} else {
$result[] = array_shift($right);
}
}
while (count($left))
$result[] = array_shift($left);
while (count($right))
$result[] = array_shift($right);
return $result;
}
C
实例
int min(int x, int y) {
return x for (seg = 1; seg for (start = 0; start while (start1 while (start1 while (start2 if (a != arr) {
int i;
for (i = 0; i 递归版：
实例
void merge_sort_recursive(int arr[], int reg[], 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, reg, start1, end1);
merge_sort_recursive(arr, reg, start2, end2);
int k = start;
while (start1 while (start1 while (start2 for (k = start; k C++
迭代版：
实例
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"(for (int seg = 1; seg for (int start = 0; start while (start1 while (start1 while (start2 if (a != arr) { for (int i = 0; i 递归版：
实例
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 if (LeftSubArray[idxLeft] else { Array[i] = RightSubArray[idxRight]; idxRight++; } }}void MergeSort(vector &Array, int front, int end) { if (front >= end) return; int mid = (front + end) / 2; MergeSort(Array, front, mid); MergeSort(Array, mid + 1, end); Merge(Array, front, mid, end);}
C#
实例
public static List sort(List lst) {
if (lst.Count 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 for (int j = mid; j 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] else {
temp.Add(right[0]);
right.RemoveAt(0);
}
}
if (left.Count > 0) {
for (int i = 0; i if (right.Count > 0) {
for (int i = 0; i return temp;
}
Ruby
实例
def merge list
return list if list.size # Merge
lambda { |left, right|
final = []
until left.empty? or right.empty?
final if left.first else right.shift end
end
final + left + right
}.call merge(list[0...pivot]), merge(list[pivot..-1])
end
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