十大经典排序算法-希尔排序与归并排序
1、希尔排序
希尔排序，也称递减增量排序算法，是插入排序的一种更高效的改进版本。但希尔排序是非稳定排序算法。
希尔排序是基于插入排序的以下两点性质而提出改进方法的：
- 插入排序在对几乎已经排好序的数据操作时，效率高，即可以达到线性排序的效率；
- 但插入排序一般来说是低效的，因为插入排序每次只能将数据移动一位；
希尔排序的基本思想是：先将整个待排序的记录序列分割成为若干子序列分别进行直接插入排序，待整个序列中的记录"基本有序"时，再对全体记录进行依次直接插入排序。
1. 算法步骤
选择一个增量序列 t1，t2，……，tk，其中 ti > tj, tk = 1；
按增量序列个数 k，对序列进行 k 趟排序；
每趟排序，根据对应的增量 ti，将待排序列分割成若干长度为 m 的子序列，分别对各子表进行直接插入排序。仅增量因子为 1 时，整个序列作为一个表来处理，表长度即为整个序列的长度。
2. 示意图
代码实现
JavaScript
实例
function shellSort(arr) {
var len = arr.length,temp,gap = 1;
while(gap < len/3) {
*//动态定义间隔序列*
gap =gap*3+1;
}
for (gap; gap > 0; gap = Math.floor(gap/3)) {
for (var i = gap; i < len; i++) {
temp = arr[i];
for (var j = i-gap; j >= 0 && arr[j] > temp; j-=gap) { arr[j+gap] = arr[j];
}
arr[j+gap] = temp;
}
}
return arr;
}
Python
实例
def shellSort(arr):
import math gap=1
while(gap < len(arr)/3):
gap = gap*3+1
while gap > 0:
for i in range(gap,len(arr)):
temp = arr[i]
j = i-gap
while j >=0 and arr[j] > temp:
arr[j+gap]=arr[j]
j-=gap
arr[j+gap] = temp
gap = math.floor(gap/3)
return arr
Go
实例
func shellSort(arr []int) []int {
length := len(arr)
gap := 1
for gap < length/3 {
gap = gap*3 + 1
}
for gap > 0 {
for i := gap; i < length; i++ {
temp := arr[i]
j := i - gap
for j >= 0 && arr[j] > temp {
arr[j+gap] = arr[j]
j -= gap
}
arr[j+gap] = temp
}
gap = gap / 3
}
return arr
}
Java
实例
public static void shellSort(int[] arr) {
int length = arr.length;
int temp;
for (int step = length / 2; step >= 1; step /= 2) {
for (int i = step; i < length; i++) {
temp = arr[i];
int j = i - step;
while (j >= 0 && arr[j] > temp) {
arr[j + step] = arr[j];
j -= step;
}
arr[j + step] = temp;
}
}
}
PHP
实例
function shellSort($arr){
$len = count($arr);
$temp = 0;
$gap = 1;
while($gap < $len / 3) {
$gap = $gap * 3 + 1;
}
for ($gap; $gap > 0; $gap = floor($gap / 3)) {
for ($i = $gap; $i < $len; $i++) {
$temp = $arr[$i];
for ($j = $i - $gap; $j >= 0 && $arr[$j] > $temp; $j -= $gap) {
$arr[$j+$gap] = $arr[$j];
}
$arr[$j+$gap] = $temp;
}
}
return $arr;
}
C
实例
void shell_sort(int arr[], int len) {
int gap, i, j;
int temp;
for (gap = len >> 1; gap > 0; gap >>= 1)
for (i = gap; i < len; i++) {
temp = arr[i];
for (j = i - gap; j >= 0 && arr[j] > temp; j -= gap)
arr[j + gap] = arr[j];
arr[j + gap] = temp;
}
}
C++
实例
template<typename T>
void shell_sort(T array[], int length) {
int h = 1;
while (h < length / 3) {
h = 3 * h + 1;
}
while (h >= 1) {
for (int i = h; i < length; i++) {
for (int j = i; j >= h && array[j] < array[j - h]; j -= h) {
std::swap(array[j], array[j - h]);
}
}
h = h / 3;
}
}
2、归并排序
归并排序（Merge sort）是建立在归并操作上的一种有效的排序算法。该算法是采用分治法（Divide and Conquer）的一个非常典型的应用。
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。
2. 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。
3. 示意图
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 < 2) {
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] <= right[0]) {
result.push(left.shift());
} 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)<2):
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] <= right[0]:
result.append(left.pop(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 < 2 {
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] <= right[0] {
result = append(result, left[0])
left = left[1:]
} 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 < 2) {
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] <= right[0]) {
result[i++] = left[0];
left = Arrays.copyOfRange(left, 1, left.length);
} 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 < 2) {
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 < 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：
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 <= end1 && start2 <= end2)
reg[k++] = arr[start1] < arr[start2] ? arr[start1++] : arr[start2++];
while (start1 <= end1)
reg[k++] = arr[start1++];
while (start2 <= end2)
reg[k++] = arr[start2++];
for (k = start; k <= end; k++)
arr[k] = reg[k];
}
void merge_sort(int arr[], const int len) {
int reg[len];
merge_sort_recursive(arr, reg, 0, len - 1);
}
C++
迭代版：
实例
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;
}
递归版：
实例
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) / 2;
MergeSort(Array, front, mid);
MergeSort(Array, mid + 1, end);
Merge(Array, front, mid, end);
}
C#
实例
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
实例
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