这是mergesort的正确实现吗？

Question

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，但我真的无法弄清楚另一种方法。请告诉我什么是错的。

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 
} 

Answer 1

您不应该需要创建多个临时数组才能执行mergesort。你做错了什么是复制数组传递给递归调用;你应该改为传递原始数组。

查看JDK中mergesort的实现可能是有益的 - 请参阅Arrays.java的1146行。

Answer 2

这是代码，它分配一个等于顶层输入大小的单个数组，并将其重用于所有递归。在一百万个整数中，我的机器需要大约300毫秒，Java库排序需要230毫秒。好吧没有调整努力，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]); 
}