我必须实现一个函数，给定一个0和1的矩阵，它返回一个包含1的坐标的新矩阵。例如：如果矩阵是3x3并且输出是：
1 0 1
0 1 1
0 0 1
新矩阵将为5x2，输出为：
0 0
0 2
1 1
1 2
2 2
一些忠告？我的方法是这样的：
int matrix[3][3];
for (int i = 0; i < 3; i++){
for (int j = 0; j < 3; j++){
if (matrix[i][i] == 1){
//Code i need
}
}
}
解决方案实际上取决于要求：
当然，其他解决方案是使用动态分配（使用malloc等）。
选项1和2的简化版本如下所示：
#include <stdio.h>
#include <stdlib.h>
int main(void) {
int matrix[3][3];
matrix[0][0] = 1;
matrix[0][1] = 0;
matrix[0][2] = 1;
matrix[1][0] = 0;
matrix[1][1] = 1;
matrix[1][2] = 1;
matrix[2][0] = 0;
matrix[2][1] = 0;
matrix[2][2] = 1;
//Solution 1 - If allowed to allocate matrix with size more than the result,
//i.e. if input is 3x3 matrix, then the maximum size of result matrix is 9 x 2
int resultMatrix1[9][2];
int usedCount1=0;
for (int i = 0; i < 3; i++){
for (int j = 0; j < 3; j++) {
if (matrix[i][j] == 1) {
resultMatrix1[usedCount1][0] = i;
resultMatrix1[usedCount1][1] = j;
usedCount1++;
} //end if
} //end for
} //end for
//Print the result
printf("\nSolution 1\n");
for (int i = 0; i < usedCount1; i++){
printf("%d %d\n", resultMatrix1[i][0], resultMatrix1[i][1]);
} //end for
//Solution 2 - strictly allocate matrix with size equal to the result.
//Without using dynamic allocation, meaning we need to have two passes.
//1st pass is to count the element which satisfy the criteria
int usedCount2=0;
for (int i = 0; i < 3; i++){
for (int j = 0; j < 3; j++) {
if (matrix[i][j] == 1) {
usedCount2++;
} //end if
} //end for
} //end for
int resultMatrix2[usedCount2][2]; //allocate the right size
int idx=0;
//2nd pass is to fill in the result matrix
for (int i = 0; i < 3; i++){
for (int j = 0; j < 3; j++) {
if (matrix[i][j] == 1) {
resultMatrix2[idx][0] = i;
resultMatrix2[idx][1] = j;
idx++;
} //end if
} //end for
} //end for
//Print the result
printf("\nSolution 2\n");
for (int i = 0; i < usedCount2; i++){
printf("%d %d\n", resultMatrix2[i][0], resultMatrix2[i][1]);
} //end for
return 0;
}
两种解决方案的结果都相同：
Solution 1
0 0
0 2
1 1
1 2
2 2
Solution 2
0 0
0 2
1 1
1 2
2 2
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