导读 | 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。
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 < 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// 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&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) / 2; MergeSort(Array, front, mid); MergeSort(Array, mid + 1, end); Merge(Array, front, mid, end); }
C#
实例
public static Listsort(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
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
原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/125350.html