Print matrix clockwise
Enter a matrix and print out each number in clockwise order from the outside to the inside.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12, 11,10,9,5,6,7]
limit:
0 <= matrix.length <= 100
0 <= matrix[i].length <= 100
solution
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* spiralOrder(int** matrix, int matrixSize, int* matrixColSize, int* returnSize){
// 处理初始化数据
*returnSize = 0;
if(matrixSize == 0 || *matrixColSize == 0)
return 0;
//真tnn的对简单这俩字有什么误会
int left = 0,top = 0;
int bottom = matrixSize - 1, right = *matrixColSize - 1;
int * p = malloc(sizeof(int)*101*101);
int * res = p;
// 按层输出
// 当在这个范围
while(left <= right && top <= bottom){
// 先一直向右
// (top, left) -> (top, right)
int i;
for(i = left;i <= right;i++){
*p = matrix[top][i];
p++;
(*returnSize)++;
}
if(*returnSize == (*matrixColSize)*matrixSize)
break;
// 向下
// (top+1, right) -> (bottom,right)
for(i = top+1; i <= bottom; i++){
*p = matrix[i][right];
p++;
(*returnSize)++;
}
if(*returnSize == (*matrixColSize)*matrixSize)
break;
// 向左
// 范围：需要倒序 (left + 1, bottom) -> (right, bottom)
for(i = right-1;i >= left;i--){
*p = matrix[bottom][i];
p++;
(*returnSize)++;
}
if(*returnSize == (*matrixColSize)*matrixSize)
break;
// 向上
// (bottom,left) -> (top+1, left)
for(i = bottom-1; i >= top+1; i--){
*p = matrix[i][left];
p++;
(*returnSize)++;
}
if(*returnSize == (*matrixColSize)*matrixSize)
break;
// left++;top++;
// bottom--;right--;
left++;top++;bottom--;right--;
}
return res;
}