有人可以告诉我为什么我在分拣后得到垃圾价值?
初始调用是(A,0,n)
,其中n
是数组的大小?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);
}
}
答案 0 :(得分:0)
你的逻辑中有很多动作难以阅读。很难区分错误的位置或这里有一些意图。但在这里我是如何通过修复看到的。请参阅内联评论。
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;
}