给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。

提示：

 - m == matrix.length - n == matrix[i].length - 1 <= m, n <= 10- -100 <= matrix[i][j] <= 100

题解1 循环（4个标志——根据顺时针）

class Solution {
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {const int row = matrix.size();const int column = matrix[0].size();vector<int> res;int i(0), j (0), startR(0), endR(column-1), startC(0), endC(row-1);while(startC <= endC){i = startC;j = startR;if(j <= endR){while(j <= endR)// i = startCres.push_back(matrix[i][j++]);startC ++;i ++;}else break;if(i <= endC){j = endR;while(i <= endC)// j = endRres.push_back(matrix[i++][j]);endR --;j --;}else break;if(j >= startR){i = endC;while(j >= startR)// i = endCres.push_back(matrix[i][j--]);endC --;i --;}else break;if(i >= startC){j = startR;while(i >= startC)// j = startRres.push_back(matrix[i--][j]);startR ++;}else break;}return res;}
};

题解2 方向

class Solution {
private:
// 向右、向下、向左、向上static constexpr int directions[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {if (matrix.size() == 0 || matrix[0].size() == 0) {return {};}int rows = matrix.size(), columns = matrix[0].size();vector<vector<bool>> visited(rows, vector<bool>(columns));int total = rows * columns;vector<int> order(total);int row = 0, column = 0;int directionIndex = 0;// 终止条件是 元素数目for (int i = 0; i < total; i++) {order[i] = matrix[row][column];visited[row][column] = true;int nextRow = row + directions[directionIndex][0], nextColumn = column + directions[directionIndex][1];if (nextRow < 0 || nextRow >= rows || nextColumn < 0 || nextColumn >= columns || visited[nextRow][nextColumn]) {directionIndex = (directionIndex + 1) % 4;}row += directions[directionIndex][0];column += directions[directionIndex][1];}return order;}
};