题解一：

        模拟：定义四个边界，指针按右下左上的顺序遍历，每遍历一条边，边界就减一，并且在某个方向没有可以遍历的数时直接返回。

import java.util.ArrayList;
import java.util.List;class Solution {public List<Integer> spiralOrder(int[][] matrix) {List<Integer> list = new ArrayList<>();int down = matrix.length - 1;//下边界int right = matrix[0].length - 1;//右边界int up = 0;//上边界int left = 0;//左边界boolean flag;while (true) {flag = false;for (int i = up, j = left; j <= right; j++) {//向右list.add(matrix[i][j]);flag = true;}if (!flag) return list;up++;flag = false;for (int i = up, j = right; i <= down; i++) {//向下list.add(matrix[i][j]);flag = true;}if (!flag) return list;right--;flag = false;for (int i = down, j = right; j >= left; j--) {//向左list.add(matrix[i][j]);flag = true;}if (!flag) return list;down--;flag = false;for (int i = down, j = left; i >= up; i--) {//向上list.add(matrix[i][j]);flag = true;}if (!flag) return list;left++;}}
}