嗨,我正在编写以螺旋顺序打印 2D 数组的程序。但是我subscripted value is neither array nor pointer nor vector
收到以下错误。
这是我的代码。
int *spiralOrder(const int** A, int n11, int n12, int *length_of_array) {
*length_of_array = n11 * n12; // length of result array
int *result = (int *)malloc(*length_of_array * sizeof(int));
int top = 0, bottom = n11 - 1, left = 0, right = n12 - 1;
int d = 0, j = 0, k = 0;
while (top <= bottom - 1 && left <= right - 1) {
int i;
if (d == 0) { //left to right
for (i = 0; i < right; i++) {
result[j++][k++] = A[top][i];
}
top++;
d = 1;
} else
if (d == 1) { //top to bottom
for (i = 0; i < bottom; i++) {
result[j++][k++] = A[i][right];
}
right--;
d = 2;
} else
if (d == 2) { //bottom right to left
for (i = right; i >= left; i--) {
result[j++][k++] = A[bottom][i];
}
bottom--;
d = 3;
} else
if (d == 3) { //bottom left to top
for (i = bottom; i >= top; i--) {
result[j++][k++] = A[i][left];
}
left++;
d = 0;
}
}
return result;
}
我想将结果保存在数组result
。我看到了这些错误的答案,但其中大多数都在处理一维数组。有人可以帮忙吗?
你malloc
ed result
作为一个一维数组,你把它作为一个 bidemsnionnal 数组。
假设您有两个变量row_number
数组大小col_number
,您应该像这样分配result
(并且不要忘记free()
):
int **result;
result = (int**) malloc(row_number * sizeof(int*));
for(int i=0; i<row_number; i++)
{
result[i] = (int*) malloc(col_number * sizeof(int));
}
首先,您将结果创建为 int*,但将其用作 int**。
其次,如果您打算不更改 2D 数组,请使用 int** const 而不是 const int**。
最后,您没有正确实现逻辑,等待
只有在无法实现逻辑时才进一步阅读。这是正确的实现:http://ideone.com/DfUD2T
int *spiralOrder(int** const A, int n11, int n12, int *length_of_array) {
*length_of_array = n11 * n12; // length of result array
int *result = (int *)malloc(*length_of_array * sizeof(int));
int top = 0, bottom = n11 - 1, left = 0, right = n12 - 1;
int j = 0, k = 0;
while (top <= bottom - 1 || left <= right - 1) {
int i;
//left to right
for (i = left; i <= right; i++) {
result[j++] = A[top][i];
}
top++;
//top to bottom
for (i = top; i <= bottom; i++) {
result[j++] = A[i][right];
}
right--;
//bottom right to left
for (i = right; i >= left; i--) {
result[j++] = A[bottom][i];
}
bottom--;
//bottom left to top
for (i = bottom; i >= top; i--) {
result[j++] = A[i][left];
}
left++;
}
return result;
}