“对于程序员而言，刷了Leetcode不一定能拿offer，但是不刷肯定拿不到offer。”

Lusion鲁迅

剑指 Offer 29. 顺时针打印矩阵

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
注意：本题与主站 54 题相同：https://leetcode-cn.com/problems/spiral-matrix/

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Solution

将矩阵按层考虑，每次顺时针循环会遍历top、right、bottom、left，控制好边界即可。

class Solution{
  public int[] spiralOrder(int[][] matrix) {
    if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0)
      return new int[0];
    int left = 0, right = matrix[0].length - 1, top = 0, bottom = matrix.length - 1;
    int[] res = new int[(right+1)*(bottom+1)];
    int x = 0;
    while (true) {
      for (int i = left; i <= right; i++) res[x++] = matrix[top][i];
      if (++top > bottom) break;
      for (int i = top; i <= bottom; i++) res[x++] = matrix[i][right];
      if (left > --right) break;
      for (int i = right; i >= left; i--) res[x++] = matrix[bottom][i];
      if (top > --bottom) break;
      for (int i = bottom; i >= top; i--) res[x++] = matrix[i][left];
      if (++left > right) break;
    }
    return res;
  }
}