题意:

给一个二维数组,按螺旋顺序将其放入一个一维数组里。

举例:

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

思路:

考察逻辑能力,纯粹的找规律,先自左向右取最上面一行,再自上而下取最右,自右向左取最下,自下而上取最左。每次都先检查释放已经取完。

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

	int bot = matrix.size() - 1, right = matrix[0].size() - 1, top = 0, left = 0;
	int rowNum = matrix.size(), colNum = matrix[0].size();
	
	vector<int> res;
	while (res.size() < rowNum*colNum) {
		for (int col = left; col <= right; ++col)
			res.push_back(matrix[top][col]);
		top++;

		if (res.size() < rowNum*colNum) {
			for (int row = top; row <= bot; ++row)
				res.push_back(matrix[row][right]);
			right--;
		}

		if (res.size() < rowNum*colNum) {
			for (int col = right; col >= left; --col)
				res.push_back(matrix[bot][col]);
			bot--;
		}

		if (res.size() < rowNum*colNum) {
			for (int row = bot; row >= top; --row)
				res.push_back(matrix[row][left]);
			left++;
		}	
	}
	return res;
}