54. Spiral Matrix(中等)

Given a matrix of m x n elements (m rows, n columns), 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].

下面的代码是这样办的： 1 2 3, 6 9, 8 7, 4, 5.

题意为旋转输出矩阵元素．
人家想法： travel路线: right -> down -> left -> up 循环直到违反循环条件(rb <= re && cb <= ce)为止．
重点是控制 row_begin, row_end, col_begin and col_end. 注意循环条件，以及 left 和 up 时的条件判断．
下面代码展现的思想特清楚．

人个想法，自个代码:
\(O(n^2)\) time, \(O(1)\) extra space.

vector<int> spiralOrder(vector<vector<int>>& A) {
	vector<int> res;
	if (A.size() == 0) return res;

	const int m = A.size(), n = A[0].size();
	int rb = 0, re = m - 1, cb = 0, ce = n - 1;

	while (rb <= re && cb <= ce) {
		// right travel
		for (int j = cb; j <= ce; j++)
			res.push_back(A[rb][j]);
		rb++;

		// down travel
		for (int j = rb; j <= re; j++)
			res.push_back(A[j][ce]);
		ce--;

		// left travel
		if (rb <= re) {
			for (int j = ce; j >= cb; j--)
				res.push_back(A[re][j]);
		}
		re--;

		// up travel
		if (cb <= ce) {
			for (int j = re; j >= rb; j--)
				res.push_back(A[j][cb]);
		}
		cb++;
	}
	return res;
}