Question

我已经苦苦挣扎了几天，想出了遵循下面的伪代码的代码来计算未排序的排列数列表的反转次数。我需要算法在O（nlogn）时间内运行，但我只能想到O（n ^ 2logn）时间内的解决方案。

更具体地说，我想知道如何通过不使用嵌套for循环来加快第二步。我知道还有其他有效的算法（即merge-sort）可以工作，但我需要按照伪代码的步骤进行操作。

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：算法应返回列表中的反转次数，即标量。伪代码要求一个数组，但最后将数组的元素相加以计算反转计数。

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

Answer 1

这是一个非常聪明的算法 - 在每次迭代中，它计算将被除法除去的反转......尽管不需要为B使用数组，因为所有你做它是添加到元素然后总结它们。你可以保留一笔运行金额。

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中所有奇数值的计数，如下所示：

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

Answer 2

在Java中不使用合并排序：

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;
    }

不使用合并排序的反演计数算法（c ++）

2 个答案: