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]
.
Analysis:
We could solve this problem straight forward. Output is element in order: up->right->down->left layer from layer.
every time we ouput its most outside rectangle, use (x1,y1) (x2,y2) represent
it can be achieved in while loop, the codition is x1<=x2 && y1<=y2
Java
public List<Integer> spiralOrder(int[][] matrix) {
ArrayList<Integer> result = new ArrayList<Integer>();
int x1 = 0;
int y1 = 0;
int x2 = matrix.length-1;
if(x2<0) return result;
int y2 = matrix[0].length-1;
while(x1<=x2 && y1<=y2){
for(int i=y1;i<=y2;i++){//right
result.add(matrix[x1][i]);
}
for(int j=x1+1;j<=x2;j++){//down
result.add(matrix[j][y2]);
}
if(x1!=x2){
for(int i=y2-1;i>=y1;i--){//left
result.add(matrix[x2][i]);
}
}
if(y1!=y2){//up
for(int j=x2-1;j>x1;j--){
result.add(matrix[j][y1]);
}
}
x1++; y1++; x2--;y2--;
}
return result;
}
c++
vector<int> spiralOrder(vector<vector<int> > &matrix) {
vector<int> result;
if(matrix.size() == 0) return result;
int x1 = 0;
int y1 = 0;
int x2 = matrix.size() - 1;
int y2 = matrix[0].size() - 1;
while(x1<=x2 && y1<=y2){
//up row
for(int j = y1; j<=y2; ++j){
result.push_back(matrix[x1][j]);
}
// right column
for(int i = x1+1; i<=x2; ++i){
result.push_back(matrix[i][y2]);
}
if(x2 !=x1){
// bottom row
for(int j = y2-1; j>=y1; --j){
result.push_back(matrix[x2][j]);
}}
if(y2 != y1){
// left column
for(int i = x2-1; i>x1; --i){
result.push_back(matrix[i][y1]);
}}
x1++, y1++, x2--, y2--;
}
return result;
}