C++ 循环置换
我正在尝试编写一个排列数组的函数C++ 循环置换,c++,C++,我正在尝试编写一个排列数组的函数 #include <iostream> #include "print.h" #include "random.h" #include <memory> int* permute_by_cycle(int A[], int size) { int dest; int* C = new int[size]; int last = size - 1; int offset = random(0, last);
#include <iostream>
#include "print.h"
#include "random.h"
#include <memory>
int* permute_by_cycle(int A[], int size)
{
int dest;
int* C = new int[size];
int last = size - 1;
int offset = random(0, last);
std::cout << "offset = " << offset << std::endl;
for(int i = 0; i < size; i++) {
dest = i + offset;
//std::cout << "dest = " << dest << "\tlast = " << last << std::endl;
if(dest > last)
dest -= last;
C[dest] = A[i];
}
return C;
}
int main()
{
int size = 18;
int A[size];
//int* B = new int[size];
fill(A,size);
print(A,size);
int* B = permute_by_cycle(A, size);
print(B,size);
delete [] B;
return 0;
}
但是,只要偏移量
大于零,其中一个元素就不会被A[i]
替换,只剩下默认的初始化值。我似乎不知道问题出在哪里。代码中的fill
和print
函数只是用随机元素填充数组并打印数组的函数
#include <iostream>
#include "print.h"
#include "random.h"
#include <memory>
int* permute_by_cycle(int A[], int size)
{
int dest;
int* C = new int[size];
int last = size - 1;
int offset = random(0, last);
std::cout << "offset = " << offset << std::endl;
for(int i = 0; i < size; i++) {
dest = i + offset;
//std::cout << "dest = " << dest << "\tlast = " << last << std::endl;
if(dest > last)
dest -= last;
C[dest] = A[i];
}
return C;
}
int main()
{
int size = 18;
int A[size];
//int* B = new int[size];
fill(A,size);
print(A,size);
int* B = permute_by_cycle(A, size);
print(B,size);
delete [] B;
return 0;
}
替换
if(dest > last)
dest -= last;
与:
if(dest > last)
dest -= size;
您应该使用模
操作符来管理循环访问。我的首选版本是:
for(int i = 0; i < size; i++) {
C[(i+offset)%size] = A[i];
}
for(int i=0;i
这样,您可以删除
dest
和所有相关的易出错行:)您是否尝试过std::next_permutation
?可能有用(除非你只是想实现算法)。我曾试图实现该算法,但我也会看一看。