反转计数算法无需使用合并排序(C ）

Inversion Count Algorithm without using Merge Sort (c++)

本文关键字：合并 排序 算法      更新时间：2023-10-16

我一直在努力几天，以提出以下伪代码以计算未分类置换数字列表的倒置数量的代码。我需要在O(nlogn(时间运行的算法，但我只能想到O(n^2logn(时间的解决方案。

更具体地说，我想知道如何通过不使用嵌套循环来加快第二步。我知道还有其他有效的算法(即合并选项(可以起作用，但我需要遵循psoudo代码的步骤。

Instance: An array A[1] . . . A[n], a permutation of n numbers 1, . . . , n
Question: Calculate vector B[j] = |{A[i] : j > i and A[i] > A[j]}| (the same as 
          B[j] = |{i : j > i and A[i] > A[j]}|) B[j] is the number of element 
          larger than A[j] to the left of A[j] in t the array A. In other words, 
          the sum of the elements in B is equal to the number of inversions in 
          the permutation A[1] . . . A[n].
(1) Initialize B[i] to 0.
(2) For each even A[j] find elements with indices smaller than j that are by one larger
than A[j]: increase B[j] by the number of such elements;
(3) Divide each A[i] by 2 (in the integer sense);
(4) Stop when all A[i] are 0.

以下是我到目前为止提出的代码：

long long numInversions = 0;     
// number of elements that are zero in the array
unsigned int zeros = 0;
do {
   // solution will need to replace this nested
   // for loop so it is O(n) not O(n^2)
   for (int i = 0; i < permNumber; i++){
           // checks if element is even
           if(array[i] % 2 == 0){
                  for (int j = i; j >= 0; j--){
                         if (array[j] == array[i] + 1){
                                numInversions++;
                         }
                 }
           }
      }
     // resets value of zeros for each pass
     zeros = 0;
     for (int k = 0; k < permNumber; k++){
             array[k] = array[k] / 2;
             if (array[k] == 0)
                  zeros++;

      }
} while(zeros != permNumber);

注意：该算法应返回列表中的反转数，标量。伪代码要求一个数组，但最终将数组的元素求和以计算反转计数。

Example: Consider a permutation (2, 3, 6, 1, 3, 5) with six inversions. The 
above algorithm works as follows:
2 4 6 1 3 5        (no pairs)                                  ÷2
1 2 3 0 1 2 1 =    0: one '1' to left, 2: one 3 to left        ÷2
0 1 1 0 0 1 1 =    0: two '1's to left, 0: two '1's to left    ÷2
0 0 0 0 0 0        total: 6 pairs

这是一种非常聪明的算法 - 在每种迭代中，它都会计算两分师将删除的反转...尽管不需要为B使用数组，因为您可以将其添加到元素中，然后将它们总结。您只能保留一个运行总和。

无论如何...为了加快步骤(2(，您可以使用另一个数组C[v]来记住A中所有奇数的计数，例如：

Step 2:
   Initialize all C[v] to 0
   For i = 1 to n:  //0 to n-1 if you're using 0-based arrays
       if A[i] is even then:
           B[i] += C[A[i]+1]
       else:
           C[A[i]] += 1

不使用Merge Sort，在Java中：

public int binarySearch(ArrayList<Integer> A, int s, int e, int elem){
        // finds the position at which elem can be inserted to maintain the sorted order in A
        if (s >= e){
            return elem >= A.get(s) ? s+1 : s;
        }
        int mid = (s+e)/2;
        if (elem == A.get(mid)) return mid+1;
        if (elem < A.get(mid)) return binarySearch(A, s, mid-1, elem); 
        return binarySearch(A, mid+1, e, elem);
    }
    public int binarySearchLast(ArrayList<Integer> A, int s, int e, int elem){
        // finds the index of first element greater than "elem" in list A
        if (s >= e) return elem < A.get(s) ? s : s+1;
        int mid = (s+e)/2;
        if (elem < A.get(mid)) return binarySearchLast(A, s, mid, elem); 
        return binarySearchLast(A, mid+1, e, elem);
    }
    public int countInversions(ArrayList<Integer> A) {
        int cnt = 0;
        ArrayList<Integer> B = new ArrayList<>();
        B.add(A.get(0));
        for (int i = 1; i < A.size(); i++){
            int idx = binarySearch(B, 0, B.size()-1, A.get(i));
            B.add(idx, A.get(i));
            idx = binarySearchLast(B, 0, B.size()-1, A.get(i));
            cnt = cnt + B.size() - idx;
        }
        return cnt;
    }
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