剑指Offer-顺时针打印矩阵

📝题目

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

拆解成多层（多个“口”字型路径），内层循环按“上边->右边->下边->左边”逐边遍历，注意检查上下边界或左右边界是否已越界，及时终止循环。

📝题解

vector<int> spiralOrder(vector<vector<int>>& matrix) {
    vector<int> res;
    int row = matrix.size();
    if (!row)    return res;
    int col = matrix[0].size();
    
    for (int layer = 0; layer * 2 < row; layer++) {
        if (layer*2+1 > row)   break;
        //遍历“上边”
        for (int j = layer; j < col - layer; j++) {
            res.push_back(matrix[layer][j]);
        }
        if (layer*2+1 == row)   break;
        
        if (layer*2+1 > col)   break;
        //遍历“右边”
        for (int i = layer+1; i < row - layer; i++) {
            res.push_back(matrix[i][col-1-layer]);
        }
        if (layer*2+1 == col)   break;
        
        //遍历“下边”
        for (int j = col-layer-2; j >= layer; j--) {
            res.push_back(matrix[row-1-layer][j]);
        }
        
        //遍历“左边”
        for (int i = row-layer-2; i > layer; i--) {
            res.push_back(matrix[i][layer]);
        }
    }
    return res;
}