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:
依次从四个方向判断。
时间复杂度:O(n*m)
空间复杂度:O(1)
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<int> res;
if (matrix.empty()) return res;
int row = matrix.size(), col = matrix[0].size();
int top = 0, bottom = row - 1, left = 0, right = col - 1;
while (true) {
for (int j = left; j <= right; ++j) res.push_back(matrix[top][j]);
if (++top > bottom) break;
for (int i = top; i <= bottom; ++i) res.push_back(matrix[i][right]);
if (--right < left) break;
for (int j = right; j >= left; --j) res.push_back(matrix[bottom][j]);
if (--bottom < top) break;
for (int i = bottom; i >= top; --i) res.push_back(matrix[i][left]);
if (++left > right) break;
}
return res;
}
};
原文:https://www.cnblogs.com/galaxy-hao/p/12709220.html