剑指 Offer 29 顺时针打印矩阵

package SwordOffer;
/** 
* @Description: 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]
 

限制：

0 <= matrix.length <= 100
0 <= matrix[i].length <= 100

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。 
* @Param:  
* @return:  
* @Author: lvhong
* @Date:  
* @E-mail lvhong282@163.com
*/ 
public class lab29easy {
    //辅助数组
    // public int[] spiralOrder(int[][] matrix) {
    //     if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
    //         return new int[0];
    //     }

    //     int width=matrix[0].length;
    //     int high=matrix.length;
    //     boolean [][] sup = new boolean[high][width];
    //     int length=width*high;
    //     int [] out =new int[length];
    //     int i=0,row=0,col=0;
    //     int [][] direction = {{0,1},{1,0},{0,-1},{-1,0}};
    //     int number=0;
    //     while(i<length){
    //         out[i]=matrix[row][col];
    //         sup[row][col]=true;
    //         int nextRow=row+direction[number][0];
    //         int nextCol=col+direction[number][1];
    //         if(nextRow<0||nextRow>=high||nextCol<0||nextCol>=width||sup[nextRow][nextCol]){
    //             number=(number+1)%4;
    //         }
    //         row+=direction[number][0];
    //         col+=direction[number][1];
    //         i++;
    //     }
    //     return out;
    // }

// //加边界
    //  public int[] spiralOrder(int[][] matrix) {
    //     if(matrix.length == 0) return new int[0];
    //     int l = 0, r = matrix[0].length - 1, t = 0, b = matrix.length - 1, x = 0;
    //     int[] res = new int[(r + 1) * (b + 1)];
    //     while(true) {
    //         for(int i = l; i <= r; i++) res[x++] = matrix[t][i]; // left to right.
    //         if(++t > b) break;
    //         for(int i = t; i <= b; i++) res[x++] = matrix[i][r]; // top to bottom.
    //         if(l > --r) break;
    //         for(int i = r; i >= l; i--) res[x++] = matrix[b][i]; // right to left.
    //         if(t > --b) break;
    //         for(int i = b; i >= t; i--) res[x++] = matrix[i][l]; // bottom to top.
    //         if(++l > r) break;
    //     }
    //     return res;
    // }


    // //按层模拟
    public int[] spiralOrder(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return new int[0];
        }
        int rows = matrix.length, columns = matrix[0].length;
        int[] order = new int[rows * columns];
        int index = 0;
        int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
        while (left <= right && top <= bottom) {
            for (int column = left; column <= right; column++) {
                order[index++] = matrix[top][column];
            }
            for (int row = top + 1; row <= bottom; row++) {
                order[index++] = matrix[row][right];
            }
            if (left < right && top < bottom) {
                for (int column = right - 1; column > left; column--) {
                    order[index++] = matrix[bottom][column];
                }
                for (int row = bottom; row > top; row--) {
                    order[index++] = matrix[row][left];
                }
            }
            left++;
            right--;
            top++;
            bottom--;
        }
        return order;
    }
}