【LeetCode】剑指 Offer 29. 顺时针打印矩阵

题目链接
输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。

示例1：

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]

示例2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]

跟54 类似

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if(matrix.length == 0){
            return new int[0];
        }
        int lenY = matrix.length;
        int lenX = matrix[0].length;
        int num = lenX * lenY;
        int count = 0;
        // 遍历下标
        int i = 0;
        int j = 0;
        // 循环次数
        int loop = Math.min(lenX, lenY) / 2;
        int midX = lenX / 2;
        int midY = lenY / 2;
        // 偏移量
        int offsetX = 1;
        int offsetY = 1;
        // 结果数组
        int[] result = new int[num];
        while (loop > 0){
            // 上
            while (j < lenX - offsetX){
                result[count] = matrix[i][j];
                count++;
                j++;
            }
            // 右
            while (i < lenY - offsetY){
                result[count] = matrix[i][j];
                i++;
                count++;
            }
            // 下
            while (j > offsetX - 1){
                result[count] = matrix[i][j];
                j--;
                count++;
            }
            // 左
            while (i > offsetY - 1){
                result[count] = matrix[i][j];
                i--;
                count++;
            }
            loop--;
            offsetX++;
            offsetY++;
            i++;
            j++;
            if(count == num){
                return result;
            }
        }
        if(lenX < lenY){
            for(;i < lenY - Math.min(lenX, lenY) / 2;i++){
                result[count] = matrix[i][j];
                count++;
            }
        }
        if(lenX > lenY){
            for(;j < lenX - Math.min(lenX, lenY) / 2;j++){
                result[count] = matrix[i][j];
                count++;
            }
        }
        if(lenX == lenY && lenX % 2 != 0){
            result[count] = matrix[midX][midY];
        }
        for (int k = 0;k<num;k++){
            System.out.println(result[k]);
        }
        return result;
    }
}