[LeetCode] Spiral Matrix, problem solving report

Preface The first blog of the New Year should first come to a problem-solving report, mainly because I was too lazy to take care of my New Year's Eve. My girlfriend had a fever and did not have time to write a year-end summary, I have some experience in the Problem Solving report. I will share it with you!

Here, I wish you a happy New Year and good luck.
Question 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:



You shoshould return [1, 2, 3, 6, 9, 8, 7, 4, 5].

The idea is to parse the matrix in the order of (top-> right-> bottom-> left) and consider the situation where only one row or only one column exists. Do not worry.



AC code

public class Solution {    public ArrayList<Integer> spiralOrder(int[][] matrix) {        int minX, minY, maxX, maxY, x, y;        ArrayList<Integer> list = new ArrayList<Integer>();        // special case        if (matrix == null || matrix.length == 0) {            return list;        }        // initial variable        minX = minY = 0;        maxX = matrix.length - 1;        maxY = matrix[0].length - 1;        for (; minX <= maxX && minY <= maxY; minX++, minY++, maxX--, maxY--) {            x = minX;            y = minY;            list.add(matrix[x][y]);            // only a row            if (minX == maxX) {                for (y += 1; y <= maxY; y++) {                    list.add(matrix[x][y]);                }                break;            }            // only a column            if (minY == maxY) {                for (x += 1; x <= maxX; x++) {                    list.add(matrix[x][y]);                }                break;            }            // a square            for (y += 1; y <= maxY; y++) { // top                list.add(matrix[x][y]);                if (y == maxY) break;            }            for (x += 1; x <= maxX; x++) { // right                list.add(matrix[x][y]);                if (x == maxX) break;            }            for (y -= 1; y >= minY; y--) { // bottom                list.add(matrix[x][y]);                if (y == minY) break;            }            for (x -= 1; x > minX; x--) { // left                list.add(matrix[x][y]);                if (x == minX + 1) break;            }        }        return list;    }}