Spiral Matrix 

 1 class Solution {
 2 public:
 3     vector<int> spiralOrder(vector<vector<int> > &matrix) {
 4         vector<int> res;
 5         if (matrix.empty() || matrix[0].empty()) return res;
 6         int imin = 0, imax = matrix.size()-1;
 7         int jmin = 0, jmax = matrix[0].size()-1;
 8         while (true) {
 9             for (int j = jmin; j <= jmax; ++j) 
10                 res.push_back(matrix[imin][j]);
11             if (++imin > imax) break;
12             for (int i = imin; i <= imax; ++i) 
13                 res.push_back(matrix[i][jmax]);
14             if (jmin > --jmax) break;
15             for (int j = jmax; j >= jmin; --j) 
16                 res.push_back(matrix[imax][j]);
17             if (imin > --imax) break;
18             for (int i = imax; i >= imin; --i) 
19                 res.push_back(matrix[i][jmin]);
20             if (++jmin > jmax) break;
21         }
22         return res;
23     }
24 };

 

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
You should return [1,2,3,6,9,8,7,4,5].

 

posted @ 2014-04-16 08:07  beehard  阅读(120)  评论(0)  编辑  收藏  举报