归并排序

归并排序

一个归并排序的例子:对一个随机点的链表进行排序
分類	排序算法
數據結構	数组
最差時間復雜度	$\Theta(n\log n)$
最優時間復雜度	$\Theta(n)$
平均時間復雜度	$\Theta(n\log n)$
最差空間復雜度	$\Theta(n)$
最佳算法	有时是

归并排序（Merge sort，台灣譯作：合併排序）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。

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。

算法描述

归并操作的过程如下：

1. 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
2. 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
3. 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
4. 重复步骤3直到某一指针达到序列尾
5. 将另一序列剩下的所有元素直接复制到合并序列尾

示例代碼

C#语言

        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;
        }

C++語言

//#只完成兩段之間歸併的功能#%
void Merge(int a[], int b[], int low, int mid, int high)
{
    int k = low;
    int begin1 = low;
    int end1 = mid;
    int begin2 = mid + 1;
    int end2 = high;
    while(k <= high )
    {
        if(begin1 > end1)
            b[k++] = a[begin2++];
        else if(begin2 > end2)
            b[k++] = a[begin1++];
        else
        {
            if(a[begin1] <= a[begin2])
                b[k++] = a[begin1++];
            else
                b[k++] = a[begin2++];
        }
    }
 
}
 
void MergePass(int a[], int b[], int seg, int size)
{
    int seg_start_ind = 0;
    while(seg_start_ind <= size - 2 * seg) //#size - 2 * seg的意思是滿足可兩兩歸併的最低臨界值#%
    {
        Merge(a, b, seg_start_ind, seg_start_ind + seg - 1, seg_start_ind + seg * 2 - 1);
        seg_start_ind += 2 * seg;
    }
    //#如果一段是正好可歸併的數量而另一段則少於正好可歸併的數量#%
    if(seg_start_ind + seg < size)
        Merge(a, b, seg_start_ind, seg_start_ind + seg - 1, size - 1);
    else
        for(int j = seg_start_ind; j < size; j++) //#如果只剩下一段或者更少的數量#%
            b[j] = a[j];
}
 
void MergeSort(int a[], int size)
{
    int* temp = new int[size];
    int seg = 1;
    while(seg < size)
    {
        MergePass(a, temp, seg, size);
        seg += seg;
        MergePass(temp, a, seg, size);
        seg += seg;
    }
    delete [] temp;
}
 
int main()
{
    int a[] = {3, 5, 3, 6, 4, 7, 5, 7, 4};
    MergeSort(a, sizeof(a) / sizeof(*a));
    //#輸出#%
    for(int i = 0; i < sizeof(a) / sizeof(*a); i++)
        cout << a[i] << ' ';
    cout << endl;
 
    return 0;
}

Ruby

def merge(left, right)
        final = []
        until left.empty? or right.empty?
                final << ( left.first < right.first ? left.shift : right.shift )
        end
        final + left + right
end

Java

public int[] Two_Way_Merge_Sort(int[] A, int[] B) {
        int[] C = new int[A.length + B.length];
        int k = 0;
        int i = 0;
        int j = 0;
        while(i < A.length && j < B.length) {
                if (A[i] < B[j])
                        C[k++] = A[i++];
                else
                        C[k++] = B[j++];
        }
        while (i < A.length) 
                C[k++] = A[i++];
        while (j < B.length) 
                C[k++] = B[j++];
        return C;
}
 
{{subst:Wikify/auto}}
/**
 * one good implements. <br />
 *
 * 
 */
public class Sort {
 
        public static final int CUTOFF = 11;
 
        /**
         * merge sort algorithm.
         * 
         * @param arr an array of Comparable item.
         * 1.here only use one temp array (think about it). <br />
         * 2.copy the element back after the sub merge operation. @see merge(T, T, int, int);
         * the above two points make it more efficient. <br />
         */
        @SuppressWarnings("unchecked")
        public static <T extends Comparable<? super T>> void mergeSort( T[] arr ) {
                //you may use insertionSort instead when the arr.length is not that large.
                /*if ( arr.length < CUTOFF ) {
                        insertionSort( arr );
                        return;
                }*/
 
                T[] tmpArr = (T[]) new Comparable[arr.length];
 
                mergeSort(arr, tmpArr, 0, arr.length - 1);
        }
 
        /**
         * internal method to make a recursive call to merge. <br />
         * 
         * @param arr an array of Comparable items. <br />
         * @param tmpArr temp array to placed the merged result. <br />
         * @param left left-most index of the subarray. <br />
         * @param right right-most index of the subarray. <br />
         */
        private static <T extends Comparable<? super T>> 
        void mergeSort( T[] arr, T[] tmpArr,
                        int left, int right ) {
                //recursive way
                if ( left < right ) {
                        int center = ( left + right ) / 2;
                        mergeSort(arr, tmpArr, left, center);
                        mergeSort(arr, tmpArr, center + 1, right);
                        merge(arr, tmpArr, left, center + 1, right);
                }
 
                //loop instead， not working, do it youself.
                /*
                int n = 0, j;
                while ( true ) {
                        int step = ( int ) Math.pow(2, ++n);
                        int len = step / 2;
                        int count = arr.length / step;
                        int rpos;
 
                        //previous pow(2, k) elements
                        for ( j = 0; j < count; j++ ) {
                                rpos = j + len;
                                System.out.println(j+", "+rpos);
                                merge( arr, tmpArr, j, rpos, rpos + len - 1);
                        }
 
                        //the rest elements
                        //for () ;
 
                        if ( step * 2 >= arr.length ) break; 
                }
                 */
        } 
 
        /**
         * internal method to merge the sorted halves of a subarray. <br />
         * 
         * @param arr an array of Comparable items. <br />
         * @param tmpArr temp array to placed the merged result. <br />
         * @param leftPos left-most index of the subarray. <br />
         * @param rightPos right start index of the subarray. <br />
         * @param endPos right-most index of the subarray. <br />
         */
        private static <T extends Comparable<? super T>> void merge( T[] arr, T[] tmpArr,
                        int lPos, int rPos, int rEnd ) {
                int lEnd = rPos - 1;
                int tPos = lPos;
                int leftTmp = lPos;
 
                while ( lPos <= lEnd && rPos <= rEnd  ) {
                        if ( arr[lPos].compareTo( arr[rPos] ) <= 0 )
                                tmpArr[ tPos++ ] = arr[ lPos++ ];
                        else 
                                tmpArr[ tPos++ ] = arr[ rPos++ ];
                }
 
                //copy the rest element of the left half subarray.
                while ( lPos <= lEnd ) 
                        tmpArr[ tPos++ ] = arr[ lPos++ ];
                //copy the rest elements of the right half subarray. (only one loop will be execute)
                while ( rPos <= rEnd ) 
                        tmpArr[ tPos++ ] = arr[ rPos++ ];
 
                //copy the tmpArr back cause we need to change the arr array items.
                for ( ; rEnd >= leftTmp; rEnd-- )
                        arr[rEnd] = tmpArr[rEnd];
        }
}

Java Another

public static int[] mergeSort(int[] arr){//归并排序 --递归
        if(arr.length==1){
                return arr;
        }
        int half = arr.length/2;
        int[] arr1 = new int[half];
        int[] arr2 = new int[arr.length-half];
        System.arraycopy(arr, 0, arr1, 0, arr1.length);
        System.arraycopy(arr, half, arr2, 0, arr2.length);
        arr1 = mergeSort(arr1);
        arr2 = mergeSort(arr2);
        return mergeSortSub(arr1,arr2);
}
 
private static int[] mergeSortSub(int[] arr1,int[] arr2){//归并排序子程序
        int[] result = new int[arr1.length+arr2.length];
        int i = 0;
        int j = 0;
        int k = 0;
        while(true){
                if(arr1[i] < arr2[j]){
                        result[k] = arr1[i];
                        if(++i>arr1.length-1){
                                break;
                        }
                }else{
                        result[k] = arr2[j];
                        if(++j>arr2.length-1){
                                break;
                        }
                }
                k++;
        }
        for(;i<arr1.length;i++){
                result[++k] = arr1[i];
        }
        for(;j<arr2.length;j++){
                result[++k] = arr2[j];
        }
        return result;
}

Python

def merge(l1,l2):
    final=[]
    #对l1,l2进行排序
    l1 = sorted(l1) 
    l2 = sorted(l2)
    while l1 and l2:
        if l1[0]<=l2[0]:
            final.append(l1.pop(0))
        else:
            final.append(l2.pop(0))
    return final+l1+l2

Erlang

%% 排序外壳
sort([]) -> [];
sort([E]) -> [E];
sort(L) ->
        {L1, L2} = lists:split(round(length(L)/2), L),
        merge(sort(L1), sort(L2)).
 
%% 归并操作
merge([], L2) -> L2;
merge(L1, []) -> L1;
merge([H1|T1]=L1, [H2|T2]=L2) ->
        if H1>H2 ->  [H1] ++ merge(T1, L2);
           true  ->  [H2] ++ merge(L1, T2)
        end.

归并排序

归并排序具体工作原理如下（假设序列共有n个元素）：

1. 将序列每相邻两个数字进行归并操作，形成$floor(n/2)$个序列，排序后每个序列包含两个元素
2. 将上述序列再次归并，形成$floor(n/4)$个序列，每个序列包含四个元素
3. 重复步骤2，直到所有元素排序完毕

示例代码

示例代码为C语言，输入参数中，需要排序的数组为array[],起始索引为first，终止索引为last。调用完成后，array[]中从first到last处于升序排列。

void merge_sort(int array[], unsigned int first, unsigned int last)
{
        int mid = 0;
        if(first<last)
        {
                /*mid = (first+last)/2;*/ /*注意防止溢出*/
                /*mid = first/2 + last/2;*/
                /*mid = （(first & last) + (first ^ last) >> 1);*/
                mid = ((first & last) + ((first ^ last) >> 1));    /*修正上一句优先级错误*/
                merge_sort(array, first, mid);
                merge_sort(array, mid+1,last);
                merge(array,first,mid,last);
        }
}

如果不使用遞歸，而是迭代，則C代碼為：

void merge_sort(int *list, int length){
        int i, left_min, left_max, right_min, right_max, next;
        int *tmp = (int*)malloc(sizeof(int) * length);
        if (tmp == NULL){
                fputs("Error: out of memory\n", stderr);
                abort();
        }
        for (i = 1; i < length; i *= 2)
                for (left_min = 0; left_min < length - i; left_min = right_max){ 
                        right_min = left_max = left_min + i;
                        right_max = left_max + i;
                        if (right_max > length)
                                right_max = length;
                        next = 0;
                        while (left_min < left_max && right_min < right_max)
                                tmp[next++] = list[left_min] > list[right_min] ? list[right_min++] : list[left_min++];
                        while (left_min < left_max)
                                list[--right_min] = list[--left_max];
                        while (next > 0)
                                list[--right_min] = tmp[--next];
                }
        free(tmp);
}

相应遞歸的Python语言的例子:

def mergesort(List):
    mid=int(len(List)/2)
    if len(List)<=1:return List
    return merge(mergesort(List[:mid]),mergesort(List[mid:]))

如果不使用递归，而是迭代，则python代码为：

def iter_mergesort(List):
    """mergesort的非递归版本，from DPV's <<Algorithms>> 
       http://www.cs.berkeley.edu/~vazirani/algorithms.html
    """
    Q=[]
    for i in List:Q.append([i])
    while len(Q) >1 : Q.append(merge(Q.pop(0),Q.pop(0)))
    return Q.pop()

相应的Ruby语言的例子，通过调用归并操作的merge方法，共同完成完整的归并排序功能。

def mergeSort(array)
        return array if array.size < 2
        left = array.first(array.size/2)
        right = array.last(array.size - array.size/2)
        merge(mergeSort(left), mergeSort(right))
end

相应的Javascript语言的例子，通过调用归并操作merge方法：

Array.prototype.mergeSort=function(array){
                var merge=function(left,right){
                                var final=[];
                                while (left.length  && right.length) {
                                        final.push(left[0] <= right[0] ? left.shift() : right.shift() );
                                }
                                return final.concat(left.concat(right));
                }//end of the merge
 
                if (this.length < 2) {
                        return this;
                }
                var _left=this.slice(0,parseInt(this.length/2));
                var _right=this.slice(parseInt(this.length/2));
                return merge(_left.mergeSort(),_right.mergeSort());
    }

用法如下： a=[7,7,11,19,13,14,333,134,256,2,3] a.mergeSort() 通过alert(a.mergeSort())可查看排序结果。

算法复杂度

比较操作的次数介于$(n log n)/2$和$n log n - n + 1$。 赋值操作的次数是$(2nlogn)$。 归并算法的空间复杂度为：Θ (n)

外部連結

• Dictionary of Algorithms and Data Structures: Merge sort
• Power Programming - Merge Sort
• Merge Sort Algorithm Simulation
• Mergesort For Linked Lists