Spiral Matrix @LeetCode

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].

	package leetcode;
	import java.util.ArrayList;
	/**
	* Solution: handle it layer by layer from outer to inner.
	* The most important is find the how to find the loop condition. Use level(every layer) here.
	* Every level(layer) costs two rows and two cols. So level * 2 < rows && level * 2 < cols is the condition.
	*
	* We don't if it's a square or rectangle. So just compare rows and cols together.
	* Or we could just need to compare Math.min(rows, cols)
	*
	* left -- cols-level-2 current level should be counted. cols-(level+1)-1
	* up -- current level counted in, so starts from rows-(level+1)-1, ends in i >= level+1
	*
	* Feature(make it easy to review):
	* down matrix[i][cols - level - 1] -- up matrix[i][level] --> cols-level-1 + level = cols - 1
	* right matrix[level][i] --- left matrix[rows - level - 1][i] --> level + rows-level-1 = rows -1
	*
	* @author jeffwan
	* @date Apr 2, 2014
	*/
	public class SpiralOrder {
	public ArrayList<Integer> spiralOrder(int[][] matrix) {
	ArrayList<Integer> result = new ArrayList<Integer>();
	if (matrix == null || matrix.length == 0) {
	return result;
	}
	int rows = matrix.length;
	int cols = matrix[0].length;
	int levelNum = Math.min(rows, cols);
	int level = 0;
	while (level * 2 < levelNum) {
	for (int i = level; i < cols - level; i++) {
	result.add(matrix[level][i]);
	}
	for (int i = level + 1; i < rows - level; i++) {
	result.add(matrix[i][cols - level - 1]);
	}
	// if only one row /col remains
	if (rows - 2 * level == 1 || cols - 2 * level == 1) {
	break;
	}
	for (int i = cols - level - 2; i >= level; i--) {
	result.add(matrix[rows - level - 1][i]);
	}
	for (int i = rows - level - 2; i >= level + 1; i--) {
	result.add(matrix[i][level]);
	}
	level++;
	}
	return result;
	}
	}

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