我写了一个小代码来查找两个二维数组的交集,不幸的是它不起作用,所以也许你可以帮助我..交集是,如果位置(x,y)上的两个数字都是“1”。否则应该是“0”
void intersection(int *mat, int rows, int cols) {
int rows1 = 5, cols1 = 4; // secend matrix is in function because i just need it here
int ma2[] = { 0, 0, 1, 0, 1, // 1. Zeile
0, 0, 1, 0, 1, // 2. Zeile
0, 0, 1, 1, 0, // 3. Zeile
0, 0, 1, 0, 0 // 4. Zeile
};
int i = 0;
int j = 0;
int x = 0;
int y = 0;
while (j < cols && y < cols1) { // maybe it is better with a for loop ??
j += 1;
y += 1;
while (i < rows && x < rows1) {
i += 1;
x += 1;
if (mat[j*rows+i] == 1 && ma2[y*rows1+x] == 1) {
printf("%d ", mat[j*rows+i]);
break;
} else {
printf("%d ", mat[j*rows+i]);
break;
}
}
printf("\n");
}
}
int main (void) {
int rows = 5, cols = 4; //first matrix is in main, because i need it for other functions
int ma[] = { 0, 0, 1, 0, 0, // 1. Zeile
1, 0, 0, 0, 0, // 2. Zeile
1, 0, 1, 0, 0, // 3. Zeile
0, 0, 1, 0, 0 // 4. Zeile
};
intersection(ma, rows, cols);
return 0;
}
输出应该是(在本例中):
{ 0, 0, 1, 0, 0, // 1. Zeile
0, 0, 0, 0, 0, // 2. Zeile
0, 0, 1, 0, 0, // 3. Zeile
0, 0, 1, 0, 0 // 4. Zeile
};
但我只得到一个 1 行矩阵
问候;)
最佳答案
试试这个
#define min(x,y) ((x) < (y) ? (x) : (y))
void intersection(int *mat, int rows, int cols) {
rows = min(rows, 5);//rows <--> cols
cols = min(cols, 4);
int ma2[] = { 0, 0, 1, 0, 1, // 1. Zeile
0, 0, 1, 0, 1, // 2. Zeile
0, 0, 1, 1, 0, // 3. Zeile
0, 0, 1, 0, 0 // 4. Zeile
};
int i, j;
for(i = 0; i < cols; ++i){
for(j = 0; j < rows; ++j){
//printf("%d ", mat[i*rows + j] == ma2[i*rows + j] ? mat[i*rows + j] : 0);
printf("%d ", mat[i*rows + j] & ma2[i*rows + j]);
}
printf("\n");
}
}
关于c - 查找二维数组中的交集,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26995598/