Merge Sort - 合併排序

核心：將兩個有序對數組合併成一個更大的有序數組。通常做法為遞歸排序，並將兩個不同的有序數組合併到第三個數組中。

先來看看動圖，合併排序是一種典型的分治(divide and conquer)應用。

Python

#!/usr/bin/env python


class Sort:
    def mergeSort(self, alist):
        if len(alist) <= 1:
            return alist

        mid = len(alist) / 2
        left = self.mergeSort(alist[:mid])
        print("left = " + str(left))
        right = self.mergeSort(alist[mid:])
        print("right = " + str(right))
        return self.mergeSortedArray(left, right)

    #@param A and B: sorted integer array A and B.
    #@return: A new sorted integer array
    def mergeSortedArray(self, A, B):
        sortedArray = []
        l = 0
        r = 0
        while l < len(A) and r < len(B):
            if A[l] < B[r]:
                sortedArray.append(A[l])
                l += 1
            else:
                sortedArray.append(B[r])
                r += 1
        sortedArray += A[l:]
        sortedArray += B[r:]

        return sortedArray

unsortedArray = [6, 5, 3, 1, 8, 7, 2, 4]
merge_sort = Sort()
print(merge_sort.mergeSort(unsortedArray))

原地(in-place)合併

Java

public class MergeSort {
    public static void main(String[] args) {
        int unsortedArray[] = new int[]{6, 5, 3, 1, 8, 7, 2, 4};
        mergeSort(unsortedArray);
        System.out.println("After sort: ");
        for (int item : unsortedArray) {
            System.out.print(item + " ");
        }
    }

    private static void merge(int[] array, int low, int mid, int high) {
        int[] helper = new int[array.length];
        // copy array to helper
        for (int k = low; k <= high; k++) {
            helper[k] = array[k];
        }
        // merge array[low...mid] and array[mid + 1...high]
        int i = low, j = mid + 1;
        for (int k = low; k <= high; k++) {
            // k means current location
            if (i > mid) {
            // no item in left part
                array[k] = helper[j];
                j++;
            } else if (j > high) {
            // no item in right part
                array[k] = helper[i];
                i++;
            } else if (helper[i] > helper[j]) {
            // get smaller item in the right side
                array[k] = helper[j];
                j++;
            } else {
            // get smaller item in the left side
                array[k] = helper[i];
                i++;
            }
        }
    }

    public static void sort(int[] array, int low, int high) {
        if (high <= low) return;
        int mid = low + (high - low) / 2;
        sort(array, low, mid);
        sort(array, mid + 1, high);
        merge(array, low, mid, high);
        for (int item : array) {
            System.out.print(item + " ");
        }
        System.out.println();
    }

    public static void mergeSort(int[] array) {
        sort(array, 0, array.length - 1);
    }
}

C++

void merge (vector<int>& arr, int low, int mid, int high){
    vector<int> helper(arr.size());
    for(int k = low; k <= high; k++){
        helper[k] = arr[k];
    }
    int i = low, j = mid+1;
    for(int k = low; k <= high; k++){
        if(i > mid){
            arr[k] = helper[j];
            j++;
        }
        else if(j > high){
            arr[k] = helper[i];
            i++;
        }
        else if(helper[j] > helper[i]){
            arr[k] = helper[j];
            j++;
        }
        else{
            arr[k] = helper[i];
            i++;
        }
    }
}

void mergeSort(vector<int>& arr, int low, int high){
    int mid = low + (high - low)/2;
    mergeSort(arr, low, mid);
    mergeSort(arr, mid + 1, high);
    merge(arr, low, mid, high);
}

時間複雜度為 $O(N \log N)$, 使用了等長的輔助陣列，空間複雜度為 $O(N)$。

Reference

• Mergesort - Robert Sedgewick 的大作，非常清晰。