Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return [1,2,3,6,9,8,7,4,5].
vector<int> spiralOrder(vector<vector<int> > &matrix)
{
if(matrix.size() == 0)
return vector<int>();
int x1 = 0;
int y1 = 0;
int x2 = matrix.size()-1;//
int y2 = matrix[0].size()-1;//
vector<int> ans;
while(x1 <= x2 && y1 <= y2)
{
//up row
for(int i = y1; i <= y2; ++i) ans.push_back(matrix[x1][i]);
//right column
for(int i = x1+1; i <= x2; ++i) ans.push_back(matrix[i][y2]);
//bottom row
if(x2 != x1)
for(int i = y2-1; i >= y1; --i) ans.push_back(matrix[x2][i]);
//left column
if(y1 != y2)
for(int i = x2-1; i > x1; --i) ans.push_back(matrix[i][y1]);
x1++, y1++, x2--, y2--;
}
return ans;
}