54. Spiral Matrix (Medium)
https://leetcode.com/problems/spiral-matrix/
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
Example 1:
Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: [ [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]
Solutions
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> ans =new ArrayList<>();
if (matrix.length == 0) {
return ans;
}
int row = matrix.length;
int col = matrix[0].length;
int m = Math.min(col, row);
for (int i = 0; i < (int) Math.ceil(m / 2.0); i++) {
List<Integer> tmp = spiralOrder(matrix, i);
ans.addAll(tmp);
}
return ans;
}
private List<Integer> spiralOrder(int[][] matrix, int margin) {
List<Integer> ans =new ArrayList<>();
int row = matrix.length;
int col = matrix[0].length;
// Literally, following case will never matched, just for reuse purpose
if (margin >= col || margin >= row) {
return ans;
}
// Go through top row
for (int i = margin; i < col - margin; i++) {
ans.add(matrix[margin][i]);
}
// Go through right column
for (int i = margin + 1; i < row - margin; i++) {
ans.add(matrix[i][col - margin - 1]);
}
// Go through bottom row, in case the spiral only contains one row
for (int i = col - margin - 2; i >= margin && (row - margin - 1 > margin); i--) {
ans.add(matrix[row - margin - 1][i]);
}
// Go through left column, in case the spiral only contains one column
for (int i = row - margin - 2; i > margin && (col - margin - 1 > margin); i--) {
ans.add(matrix[i][margin]);
}
return ans;
}
}