https://leetcode.com/problems/spiral-matrix/

Given a matrix ofmxnelements (mrows,ncolumns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]

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

模拟题

思路也是简单明了：

但是对于层 我们需要确定遍历的范围，用两点就足够：top-left and bottom-right （same for rectangle）

当我们没遍历完一遍 我们可以相应的更新这两点

        int si = 0;
        int sj = 0;
        int ei = m - 1;
        int ej = n - 1;
        while (si <= ei && sj <= ej) {
            // up
            for (int j = sj; j <= ej && si <= ei && sj <= ej; ++j) {
                result.add(matrix[si][j]);
            }
            si++;
            // right
            for (int i = si; i <= ei && si <= ei && sj <= ej; ++i) {
                result.add(matrix[i][ej]);
            }
            ej--;
            // down
            for (int j = ej; j >= sj && si <= ei && sj <= ej; --j) {
                result.add(matrix[ei][j]);
            }
            ei--;
            // left
            for (int i = ei; i >= si && si <= ei && sj <= ej; --i) {
                result.add(matrix[i][sj]);
            }
            sj++;
        }

但其实 我们不需要那么多check, 1st for just from while, so no need to check again, 2nd we just update si, but we checked in our for, also no need. for 3rd j >= sj also means ej >= sj.

// following already contains all checking above

        while (si <= ei && sj <= ej) {
            // up
            for (int j = sj; j <= ej; ++j) {
                result.add(matrix[si][j]);
            }
            si++;
            // right
            for (int i = si; i <= ei; ++i) {
                result.add(matrix[i][ej]);
            }
            ej--;
            // down
            for (int j = ej; j >= sj && si <= ei; --j) {
                result.add(matrix[ei][j]);
            }
            ei--;
            // left
            for (int i = ei; i >= si && sj <= ej; --i) {
                result.add(matrix[i][sj]);
            }
            sj++;
        }