所以，我一直在尝试创建合并排序算法，但由于某些原因，结果不正确。我已经尝试打印高，低和中值，以及每次更改时的数组值，我注意到某些值的数量甚至都不正确。因此，例如，如果我输入3，最终结果将包括多个3。但真正让我感到困惑的是，当我输入4个或更少的值时，这不会发生，我只注意到五个或更高的输入值。我不知道是计算机还是编译器正在改变其中的一些值，或者我的代码是否是问题。
这是我的代码：
#include <iostream>
#include <array>
int num;
int * array = new int [num];
int * sub = new int [num];
void sort (int low, int mid, int high) {
std::cout << "Low, mid, high: " << low << ", " << mid << ", " << high << ".\n";
int a,b,c,d;
a = low;
b = low;
c = mid;
while ((a < mid) && (c < high)) {
if (array[a] < array[c]) {
sub[b] = array[a];
a++;
} else {
sub[b] = array[c];
c++;
}
b++;
}
while (a == mid && c < high) {
sub[b] = array[c];
c++;
b++;
}
while (c == high && a < mid) {
sub[b] = array[a];
a++;
b++;
}
for (d = low; d < high; d++) {
array[d] = sub[d];
for (int i = 0; i < num; i++) {
std::cout << array[i];
if (i != num - 1) {
std::cout << ", ";
} else {
std::cout << ".\n";
}
}
}
std::cout << "Done!\n";
}
void split (int low, int high) {
if (low < high - 1) {
int mid = (low + high) / 2;
split(low, mid);
split(mid, high);
sort(low, mid, high);
}
}
int main()
{
std::cout << "This is a program that sorts integers.\n";
std::cout << "How many numbers would you like to sort?\n";
std::cin >> num;
std::cout << "Please type in the numbers.\n";
for (int i = 0; i < num; i++) {
std::cin >> array[i];
}
split(0, num);
std::cout << "Your sorted numbers are: ";
for (int i = 0; i < num; i++) {
std::cout << array[i];
if (i != num - 1) {
std::cout << ", ";
} else {
std::cout << ".\n";
}
}
delete[] array;
delete[] sub;
return 0;
}
我还在原始编译器和在线shell上尝试了我的代码，似乎他们为同一个测试产生了两个不同的结果：
原始编译器：
This is a program that sorts integers.
How many numbers would you like to sort?
9
Please type in the numbers.
5
9
2
0
6
4
3
1
8
Low, mid, high: 0, 1, 2.
5, 9, 2, 0, 5, 9, 3, 1, 8.
5, 9, 2, 0, 5, 9, 3, 1, 8.
Done!
Low, mid, high: 2, 3, 4.
5, 9, 0, 0, 5, 9, 0, 2, 8.
5, 9, 0, 2, 5, 9, 0, 2, 8.
Done!
Low, mid, high: 0, 2, 4.
0, 9, 0, 2, 0, 2, 5, 9, 8.
0, 2, 0, 2, 0, 2, 5, 9, 8.
0, 2, 5, 2, 0, 2, 5, 9, 8.
0, 2, 5, 9, 0, 2, 5, 9, 8.
Done!
Low, mid, high: 4, 5, 6.
0, 2, 5, 9, 0, 2, 5, 9, 0.
0, 2, 5, 9, 0, 2, 5, 9, 0.
Done!
Low, mid, high: 7, 8, 9.
0, 2, 5, 9, 0, 2, 5, 0, 0.
0, 2, 5, 9, 0, 2, 5, 0, 9.
Done!
Low, mid, high: 6, 7, 9.
0, 2, 5, 9, 0, 2, 0, 0, 9.
0, 2, 5, 9, 0, 2, 0, 5, 9.
0, 2, 5, 9, 0, 2, 0, 5, 9.
Done!
Low, mid, high: 4, 6, 9.
0, 2, 5, 9, 0, 2, 0, 5, 0.
0, 2, 5, 9, 0, 0, 0, 5, 0.
0, 2, 5, 9, 0, 0, 2, 5, 0.
0, 2, 5, 9, 0, 0, 2, 5, 0.
0, 2, 5, 9, 0, 0, 2, 5, 0.
Done!
Low, mid, high: 0, 4, 9.
0, 2, 5, 9, 0, 0, 0, 0, 0.
0, 0, 5, 9, 0, 0, 0, 0, 0.
0, 0, 0, 9, 0, 0, 0, 0, 0.
0, 0, 0, 0, 0, 0, 0, 0, 0.
0, 0, 0, 0, 0, 0, 0, 0, 0.
0, 0, 0, 0, 0, 0, 0, 0, 0.
0, 0, 0, 0, 0, 0, 2, 0, 0.
0, 0, 0, 0, 0, 0, 2, 5, 0.
0, 0, 0, 0, 0, 0, 2, 5, 9.
Done!
Your sorted numbers are: 0, 0, 0, 0, 0, 0, 2, 5, 9.
Program ended with exit code: 0
在线shell：
This is a program that sorts integers.
How many numbers would you like to sort?
9
Please type in the numbers.
5
9
2
0
6
4
3
1
8
Low, mid, high: 0, 1, 2.
5, 9, 2, 0, 6, 4, 3, 1, 5.
5, 9, 2, 0, 6, 4, 3, 1, 5.
Done!
Low, mid, high: 2, 3, 4.
5, 9, 0, 0, 6, 4, 3, 1, 5.
5, 9, 0, 2, 6, 4, 3, 1, 5.
Done!
Low, mid, high: 0, 2, 4.
0, 9, 0, 2, 6, 4, 3, 1, 0.
0, 2, 0, 2, 6, 4, 3, 1, 0.
0, 2, 5, 2, 6, 4, 3, 1, 0.
0, 2, 5, 9, 6, 4, 3, 1, 0.
Done!
Low, mid, high: 4, 5, 6.
0, 2, 5, 9, 4, 4, 3, 1, 0.
0, 2, 5, 9, 4, 6, 3, 1, 0.
Done!
Low, mid, high: 7, 8, 9.
0, 2, 5, 9, 4, 6, 3, 0, 0.
0, 2, 5, 9, 4, 6, 3, 0, 1.
Done!
Low, mid, high: 6, 7, 9.
0, 2, 5, 9, 4, 6, 0, 0, 1.
0, 2, 5, 9, 4, 6, 0, 1, 1.
0, 2, 5, 9, 4, 6, 0, 1, 3.
Done!
Low, mid, high: 4, 6, 9.
0, 2, 5, 9, 0, 6, 0, 1, 3.
0, 2, 5, 9, 0, 1, 0, 1, 3.
0, 2, 5, 9, 0, 1, 3, 1, 3.
0, 2, 5, 9, 0, 1, 3, 4, 3.
0, 2, 5, 9, 0, 1, 3, 4, 6.
Done!
Low, mid, high: 0, 4, 9.
0, 2, 5, 9, 0, 1, 3, 4, 0.
0, 0, 5, 9, 0, 1, 3, 4, 0.
0, 0, 1, 9, 0, 1, 3, 4, 0.
0, 0, 1, 2, 0, 1, 3, 4, 0.
0, 0, 1, 2, 3, 1, 3, 4, 0.
0, 0, 1, 2, 3, 4, 3, 4, 0.
0, 0, 1, 2, 3, 4, 0, 4, 0.
0, 0, 1, 2, 3, 4, 0, 5, 0.
0, 0, 1, 2, 3, 4, 0, 5, 9.
Done!
Your sorted numbers are: 0, 0, 1, 2, 3, 4, 0, 5, 9.
正如您所看到的，我使用了完全相同的数字和代码，但得到了两个完全不同的结果，因此我不知道我的代码是错误还是其他内容。感谢您提前提供任何帮助。
答案 0 :(得分：0)
由于num在代码开头声明时未定义，因为它是全局变量num自动分配给0。然后int * array = new int [num];和int * sub = new int [num];声明了num尚未由用户定义，创建了两个带有0元素的数组。尝试更改这些数组会导致未定义的行为，其中包括错误地使某些案例看起来是正确的，并且在这种情况下，使用无关紧要的术语来更改用户创建的数字列表。
要解决此问题，只能在用户定义int * array后定义int * sub和num。这也意味着函数split和sort必须包含指向数组的指针作为参数。
以下是代码的样子：
#include <iostream>
#include <array>
void sort (int *array, int *sub, int low, int mid, int high) {
int a = low;
int b = low;
int c = mid;
while ((a < mid) && (c < high)) {
if (array[a] < array[c]) {
sub[b] = array[a];
a++;
} else {
sub[b] = array[c];
c++;
}
b++;
}
while (a == mid && c < high) {
sub[b] = array[c];
c++;
b++;
}
while (c == high && a < mid) {
sub[b] = array[a];
a++;
b++;
}
for (int d = low; d < high; d++) {
array[d] = sub[d];
}
}
void split (int *array, int *sub, int low, int high) {
if (low < high - 1) {
int mid = (low + high) / 2;
split(array, sub, low, mid);
split(array, sub, mid, high);
sort(array, sub, low, mid, high);
}
}
int main()
{
std::cout << "This is a program that sorts integers.\n";
std::cout << "How many numbers would you like to sort?\n";
int num;
std::cin >> num;
int * array = new int [num];
int * sub = new int [num];
std::cout << "Please type in the numbers.\n";
for (int i = 0; i < num; i++) {
std::cin >> array[i];
}
split(array, sub, 0, num);
std::cout << "Your sorted numbers are: ";
for (int i = 0; i < num; i++) {
std::cout << array[i];
if (i != num - 1) {
std::cout << ", ";
} else {
std::cout << ".\n";
}
}
delete[] array;
delete[] sub;
return 0;
}
答案 1 :(得分：0)
您可以考虑使用自下而上合并排序。两者的例子：
自下而上：
template <typename T>
T * BottomUpMergeSort(T a[], T b[], size_t n)
{
for(size_t s = 1; s < n; s += 2) // swap in place for 1st pass
if(a[s] < a[s-1])
std::swap(a[s], a[s-1]);
for(size_t s = 2; s < n; s <<= 1){ // s = run size
size_t ee = 0; // init end index
while(ee < n){ // merge pairs of runs
size_t ll = ee; // ll = start of left run
size_t rr = ll+s; // rr = start of right run
if(rr >= n){ // if only left run
rr = n;
BottomUpCopy(a, b, ll, rr); // copy left run
break; // end of pass
}
ee = rr+s; // ee = end of right run
if(ee > n)
ee = n;
BottomUpMerge(a, b, ll, rr, ee);
}
std::swap(a, b); // swap a and b
}
return a; // return sorted array
}
template <typename T>
void BottomUpCopy(T a[], T b[], size_t ll, size_t rr)
{
while(ll < rr){ // copy left run
b[ll] = a[ll];
ll++;
}
}
template <typename T>
void BottomUpMerge(T a[], T b[], size_t ll, size_t rr, size_t ee)
{
size_t o = ll; // b[] index
size_t l = ll; // a[] left index
size_t r = rr; // a[] right index
while(1){ // merge data
if(a[l] <= a[r]){ // if a[l] <= a[r]
b[o++] = a[l++]; // copy a[l]
if(l < rr) // if not end of left run
continue; // continue (back to while)
while(r < ee){ // else copy rest of right run
b[o++] = a[r++];
}
break; // and return
} else { // else a[l] > a[r]
b[o++] = a[r++]; // copy a[r]
if(r < ee) // if not end of right run
continue; // continue (back to while)
while(l < rr){ // else copy rest of left run
b[o++] = a[l++];
}
break; // and return
}
}
}
自上而下，避免在合并后通过在两个递归函数之间交替复制数据：
template <typename T>
T * TopDownMergeSort(T a[], T b[], size_t n)
{
TopDownMergeSortAtoA(a, b, 0, n);
return a;
}
template <typename T>
void TopDownMergeSortAtoA(T a[], T b[], size_t ll, size_t ee)
{
if (ee - ll > 1) {
size_t rr = (ll + ee)>>1; // midpoint, start of right half
TopDownMergeSortAtoB(a, b, ll, rr);
TopDownMergeSortAtoB(a, b, rr, ee);
TopDownMerge(b, a, ll, rr, ee); // merge b to a
}
}
template <typename T>
void TopDownMergeSortAtoB(T a[], T b[], size_t ll, size_t ee)
{
if (ee - ll > 1) {
size_t rr = (ll + ee)>>1; //midpoint, start of right half
TopDownMergeSortAtoA(a, b, ll, rr);
TopDownMergeSortAtoA(a, b, rr, ee);
TopDownMerge(a, b, ll, rr, ee); // merge a to b
} else if ((ee - ll) == 1) {
b[ll] = a[ll];
}
}
template <typename T>
void TopDownMerge(T a[], T b[], size_t ll, size_t rr, size_t ee)
{
size_t o = ll; // b[] index
size_t l = ll; // a[] left index
size_t r = rr; // a[] right index
while(1){ // merge data
if(a[l] <= a[r]){ // if a[l] <= a[r]
b[o++] = a[l++]; // copy a[l]
if(l < rr) // if not end of left run
continue; // continue (back to while)
while(r < ee){ // else copy rest of right run
b[o++] = a[r++];
}
break; // and return
} else { // else a[l] > a[r]
b[o++] = a[r++]; // copy a[r]
if(r < ee) // if not end of right run
continue; // continue (back to while)
while(l < rr){ // else copy rest of left run
b[o++] = a[l++];
}
break; // and return
}
}
}