0

我担心为每个递归步骤创建 3 个数组可能会占用太多空间，但我真的想不出另一种方法。请告诉我它有什么问题。

public static int[] split(int [] vector){

    if(vector.length <= 1 || vector == null) 
        return vector;
    int len = vector.length;

    int[] list1 = new int[len / 2];
    // If the number of elements is odd the second list will be bigger
    int[] list2 = new int[len / 2 + (len % 2)];

    // Here we assign the elements to 2 separate lists
    for(int x = 0; x < len / 2; x++)
        list1[x] = vector[x];
    for(int j = 0, i = len / 2; j < list2.length; i++, j++)
        list2[j]=vector[i];

    // Apply the recursion, this will eventually order the lists
    list1 = split(list1);
    list2 = split(list2);

    // Here we take the 2 ordered lists and merge them into 1
    int i = 0, a = 0, b = 0;
    int[] listfinal = new int[len];
    while(i < len){
        if(a >= list1.length){
            listfinal[i] = list2[b]; 
            b++;
        } else if(b >= list2.length){
            listfinal[i] = list1[a]; 
            a++;
        } else if(list1[a] <= list2[b]){
            listfinal[i] = list1[a]; 
            a++;
        } else if(list1[a] > list2[b]){
            listfinal[i] = list2[b]; 
            b++;
        }
        i++;
    }
    return listfinal; // Return the merged and ordered list
}
4

2 回答 2

1

您不需要创建多个临时数组来进行归并排序。您做错的是复制数组以传递给递归调用；您应该改为传递原始数组。

查看 JDK 中合并排序的实现可能会提供一些信息——查看 Arrays.java 的第 1146行。

于 2012-11-25T18:04:59.083 回答
0

这是在顶层分配一个等于输入大小的单个数组并将其重新用于所有递归的代码。在一百万个整数上，这在我的机器上大约需要 300 毫秒，而 Java 库排序需要 230 毫秒。好吧，不用调整努力，我猜...

// Sort the elements of a between lo and hi inclusive.
private static void sortImpl(int [] a, int lo, int hi, int [] tmp) {

    if (hi <= lo) return;

    // Recur on sublists.
    int mid = (hi + lo) / 2;
    sortImpl(a, lo, mid, tmp);
    sortImpl(a, mid + 1, hi, tmp);

    // Move past items already in the right place.
    int t1 = lo;
    while (a[t1] < a[mid + 1]) t1++;

    // Merge sublists into result.
    int p1 = t1;
    int p2 = mid + 1;
    int i = t1;
    System.arraycopy(a, t1, tmp, t1, mid - t1 + 1);
    while (p1 <= mid)
        a[i++] = (p2 > hi || tmp[p1] < a[p2]) ? tmp[p1++] : a[p2++];
}

public static void sort(int [] a) {
    sortImpl(a, 0, a.length - 1, new int[a.length]);
}
于 2012-11-25T19:28:51.760 回答