现在的位置: 首页 > 综合 > 正文

合并(归并)排序(MergeSort)

2012年08月07日 ⁄ 综合 ⁄ 共 1838字 ⁄ 字号 评论关闭

 //用来计数,测试执行循环次数.
        int count;
        private void button1_Click(object sender, EventArgs e)
        {
           
            Random theR = new Random(new Random().Next(30000));
            List<int> theA = new List<int>();
            int n = 1024;
            for (int i = 1; i <= n; i++)
            {
                theA.Add(theR.Next(10000));
            }
            count = 0;
            int[] theB = theA.ToArray();
            MergeSort(theB, 0, n-1);
        }

        private void MergeSort(int[] A,int iS,int iE)
        {

            //相等,表示一个数,一个数认为是已排序。
            if (iS == iE)
            {
                count++;
                return;
            }

            //分左右两段排序,2分法
            int iE1 = (iS + iE) / 2;
            int iS2 = iE1 + 1;
            MergeSort(A, iS, iE1);
            MergeSort(A, iS2, iE);

            //排完左右后进行归并针对两个排好序的段(iS-iE1,iS2-iE)进行整理
             int i1 = iS, j1 = iS2;//归并初始都指向各自段得最小索引.
            //因为都存放在数组的iS->iE段中,而且从小到大,从左到右存放。
            //最大整理次数为iE-iS,但一般只要整理完其中一段后即可.
            while(true)
            {
                //左边段当前值小于等于右边段最小值,则不需要移动,左边段指针i1右移一位即可。
                if (A[i1] <= A[j1])
                {
                    count++;
                    i1++;//右移
                    //因为两个段都是排序的,左边段如果整理完毕,则左边段不再需要整理
                    if (i1 >= j1)
                    {
                        break;
                    }
                }
                else
                {
                    //如果左边段大于右边段,则需要移位处理.
                    //将j1移到i1处,原来的i1到j1-1整体右移一位.
                    int tmp = A[i1];
                    A[i1] = A[j1];
                    for (int b2 = j1; b2 > i1+1; b2--)
                    {
                        A[b2] = A[b2 - 1];
                        count++;
                    }
                    A[i1 + 1] = tmp;
                    i1++; j1++;
                }
                //因为两个段都是排序的,右边段如果整理完毕,则左边段不再需要整理
                if (j1 > iE)
                {
                    break;
                }
            }
        }

抱歉!评论已关闭.