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

思路：直接模拟顺时针遍历矩阵的情况，每次打印完一行或一列后判断是否已经遍历完。

vector<int> spiralOrder(vector<vector<int>>& matrix) {
    if (matrix.size() == 0)
        return {};
    vector<int> res;
    int top = 0, bottom = matrix.size() - 1, left = 0, right = matrix[0].size() - 1;
    while (true) {
        for (int i = left; i <= right; i++) {
            res.push_back(matrix[top][i]);
        }
        if (++top > bottom)
            break;

        for (int i = top; i <= bottom; i++) {
            res.push_back(matrix[i][right]);
        }
        if (--right < left)
            break;

        for (int i = right; i >= left; i--) {
            res.push_back(matrix[bottom][i]);
        }
        if (--bottom < top)
            break;
            
        for (int i = bottom; i >= top; i--) {
            res.push_back(matrix[i][left]);
        }
        if (++left > right)
            break;
    }
    return res;
}