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。
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]
提示:
m == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
题解
界定好边界值即可,定义四个变量,分别表示上下左右四个边界,每一次的遍历都是遇到边界就停止,依次遍历出来即可。
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
if (matrix.empty()) return {};
int l = 0, r = matrix[0].size() - 1, t = 0, b = matrix.size() - 1;
vector<int> res;
while (true) {
for (int i = l; i <= r; i++) res.push_back(matrix[t][i]); // left to right
if (++t > b) break;
for (int i = t; i <= b; i++) res.push_back(matrix[i][r]); // top to bottom
if (l > --r) break;
for (int i = r; i >= l; i--) res.push_back(matrix[b][i]); // right to left
if (t > --b) break;
for (int i = b; i >= t; i--) res.push_back(matrix[i][l]); // bottom to top
if (++l > r) break;
}
return res;
}
}