题目:
给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
示例 1:
输入:
[
[ 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:
[
[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]
c++代码,解题思路同59题:
class Solution {
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {/*思路:同59 螺旋矩阵 2*/vector<int> res;if(matrix.empty()) return res;int up = 0;int down = matrix.size() - 1; //colint left = 0;int right = matrix[0].size() - 1; //rowwhile(up <= down && left <= right){//left --> rightfor(int i = left; i <= right; i++){ res.push_back(matrix[up][i]); //添加到res}//下一步要走的方向,依次类推if(++up > down) break;//up --> downfor(int i = up; i <= down; i++){res.push_back(matrix[i][right]);}if(--right < left) break; //right --> leftfor(int i = right; i >= left; i--){res.push_back(matrix[down][i]);}if(--down < up) break; //down --> upfor(int i = down; i >= up; i--){res.push_back(matrix[i][left]);}if(++left > right) break;}return res;}
};
