Merge_sort没有哨兵

Merge_sort without sentinel

本文关键字：哨兵 sort Merge      更新时间：2023-10-16

有人可以告诉我为什么我在排序后会得到垃圾值吗？初始调用(A,0,n)数组的大小在哪里n？我想使用合并排序算法对数组进行排序，但没有哨兵值。

void merge_sort(int A[], int l, int mid, int r)
{
    int n1 = mid - l + 1;
    int n2 = r - mid;
    int L[n1], R[n2];
    for (int i = 0; i < n1; i++)
    {
        L[i] = A[i];
    }
    for (int i = 0; i <= n2; i++)
    {
        R[i] = A[i + mid + 1];
    }
    cout << endl;
    int j = 0, k = 0;
    for (int i = l; i < r; i++)
    {
        if (j == n1 || k == n2)
        {
            if (j == n1 + 1)
            {
                A[i] = R[k];
                k++;
            }
            else
            {
                A[i] = L[j];
                j++;
            }
        }
        else if (L[j] >= R[k])
        {
            A[i] = L[j];
            j++;
        }
        else
        {
            A[i] = R[k];
            k++;
        }
    }
}
void merge_divide(int A[], int l, int r)
{
    if (l < r)
    {
        int mid = (l + r) / 2;
        merge_divide(A, l, mid);
        merge_divide(A, mid + 1, r);
        merge_sort(A, l, mid, r);
    }
}

你的逻辑中有很多动作很难阅读。很难区分哪里有错误或这里有一些意图。但这就是我对修复的看法。请参阅内联注释。

void merge_sort(int A[], int l, int mid, int r) {
  int n1 = mid - l;
  int n2 = r - mid;
  // this is C, but not C++, consider using vector instead
  int L[n1], R[n2];
  for (int i = 0; i < n1; i++) {
    // need `l` here
    L[i] = A[l + i];
  }
  for (int i = 0; i < n2; i++) {
    // need to include `mid`
    R[i] = A[i + mid];
  }
  int j = 0, k = 0;
  for (int i = l; i < r; i++) {
    if (j == n1) {
      A[i] = R[k];
      k++;
    } else if (k == n2) {
      A[i] = L[j];
      j++;
    } else if (L[j] >= R[k]) {
      A[i] = L[j];
      j++;
    } else {
      A[i] = R[k];
      k++;
    }
  }
}
void merge_divide(int A[], int l, int r) {
    // need properly compute `mid` here
  int mid = r - l;
  if (mid > 1) {
    mid = l + mid / 2;
    merge_divide(A, l, mid);
    merge_divide(A, mid, r);
    merge_sort(A, l, mid, r);
  }
}
int main() {
  int A[] = {2, 4, 3, 382, 2342334, 3, 42, 234};
  int n = sizeof(A) / sizeof(int);
  merge_divide(A, 0, n);
  for (int i = 0; i < n; ++i)
    cerr << A[i] << " ";
  cout << endl;
}
相关文章：