Spiral Matrix

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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].

class Solution {public:    vector<int> spiralOrder(vector<vector<int> > &matrix)    {         vector<int> res;         if (matrix.empty()) return res;                  int rh = 0;          int rb = matrix.size() - 1;         int cl = 0;          int cr = matrix[0].size() - 1;         while (rh <= rb || cl <= cr)         { if(rh <= rb)             for (int i = cl; i <= cr; i++)                res.push_back(matrix[rh][i]); if (cl <= cr)             for (int i = rh + 1; i <= rb; i++)                res.push_back(matrix[i][cr]); if (rh < rb)             for (int i = cr - 1; i >= cl; i--)                res.push_back(matrix[rb][i]); if (cl < cr)             for (int i = rb - 1; i >= rh + 1; i--)                res.push_back(matrix[i][cl]);                          rh++; rb--; cl++;cr--;         }         return res;    }};

要点在于四个if判断，只有if里条件满足才有遍历的意义

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