按特定顺序就地转换矩阵
编写代码以特定方式转换矩阵而不使用额外空间。
Input:
1 2 3
4 5 6
7 8 9
Output:
1 6 7
2 5 8
3 4 9乍一看，这个问题似乎类似于寻找矩阵的转置。但是如果你仔细看，你会注意到输出矩阵中的每一列偶数列都有输入矩阵中对应行的元素，顺序相反。
我们强烈建议您最小化您的浏览器并首先自己尝试。
通过对输入矩阵进行一些修改，可以轻松地将问题转换为矩阵的转置。如果我们反转输入矩阵中存在的每个偶数行，我们可以使用此处给出的解决方案以所需的顺序转换矩阵，并且也无需使用任何辅助存储器。
下面是这个想法的 C++ 实现。
C
// Program for convert matrix in specific order
// using in-place matrix transpose
#include
#define HASH_SIZE 128
using namespace std;
// Non-square matrix transpose of matrix of size r x c
// and base address A
void transformMatrix(int *A, int r, int c)
{
// Invert even rows
for (int i = 1; i < r; i = i + 2)
for (int j1 = 0, j2 = c - 1; j1 < j2; j1++, j2--)
swap(*(A + i*c + j1), *(A + i*c + j2));
// Rest of the code is from below post
// http://tinyurl.com/j79j445
int size = r*c - 1;
int t; // holds element to be replaced, eventually
// becomes next element to move
int next; // location of 't' to be moved
int cycleBegin; // holds start of cycle
bitset b; // hash to mark moved elements
b.reset();
b[0] = b[size] = 1;
int i = 1; // Note that A[0] and A[size-1] won't move
while (i < size)
{
cycleBegin = i;
t = A[i];
do
{
// Input matrix [r x c]
// Output matrix 1
// i_new = (i*r)%(N-1)
next = (i*r)%size;
swap(A[next], t);
b[i] = 1;
i = next;
} while (i != cycleBegin);
// Get Next Move (what about querying
// random location?)
for (i = 1; i < size && b[i]; i++)
;
}
}
// A utility function to print a 2D array of size
// nr x nc and base address A
void Print2DArray(int *A, int nr, int nc)
{
for (int r = 0; r < nr; r++)
{
for (int c = 0; c < nc; c++)
printf("%4d", *(A + r*nc + c));
printf("\n");
}
printf("\n");
}
// Driver program to test above function
int main(void)
{
int A[][4] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12}};
int r = 3, c = 4;
cout << "Given Matrix:\n";
Print2DArray((int *)A, r, c);
transformMatrix((int *)A, r, c);
cout << "Transformed Matrix:\n";
Print2DArray((int *)A, c, r);
return 0;
} C++
// C++ Program for convert matrix in specific order
// using in-place matrix transpose
#include
#define HASH_SIZE 128
using namespace std;
// Non-square matrix transpose of matrix of size r x c
// and base address A
void transformMatrix(int *A, int r, int c)
{
// Invert even rows
for (int i = 1; i < r; i = i + 2)
for (int j1 = 0, j2 = c - 1; j1 < j2; j1++, j2--)
swap(*(A + i*c + j1), *(A + i*c + j2));
// Rest of the code is from below post
// http://tinyurl.com/j79j445
int size = r*c - 1;
int t; // holds element to be replaced, eventually
// becomes next element to move
int next; // location of 't' to be moved
int cycleBegin; // holds start of cycle
bitset b; // hash to mark moved elements
b.reset();
b[0] = b[size] = 1;
int i = 1; // Note that A[0] and A[size-1] won't move
while (i < size)
{
cycleBegin = i;
t = A[i];
do
{
// Input matrix [r x c]
// Output matrix 1
// i_new = (i*r)%(N-1)
next = (i*r)%size;
swap(A[next], t);
b[i] = 1;
i = next;
} while (i != cycleBegin);
// Get Next Move (what about querying
// random location?)
for (i = 1; i < size && b[i]; i++)
;
}
}
// A utility function to print a 2D array of size
// nr x nc and base address A
void Print2DArray(int *A, int nr, int nc)
{
for (int r = 0; r < nr; r++)
{
for (int c = 0; c < nc; c++)
{
cout< 输出：
Given Matrix:
1 2 3 4
5 6 7 8
9 10 11 12
Transformed Matrix:
1 8 9
2 7 10
3 6 11
4 5 12