【leetcode】 Spiral Matrix

Spiral Matrix

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].

题意很简单，就是顺时针螺旋打印矩阵。思路也很简单，就是把元素分别从左向右、上到下、右到左和下到上螺旋保存到一个数组中。但是需要注意细节，特别是边界条件判断。

 #include <iostream>
 #include <vector>

 using namespace std;
 class Solution {
 public:
     vector<int> spiralOrder(vector<vector<int> > &matrix) {
         vector<int> ans;
         if(matrix.empty())
             return ans;
         int begin_row = , end_row = matrix[].size() - ;//begin_row is the row number, end_row is the remaind size of each row
         int begin_col = , end_col = matrix.size() - ;//begin_col is the col number,end_col is the remaind size of each col
         while(true){
             for(int i = begin_row; i <= end_row; ++i)//left to right
                 ans.push_back(matrix[begin_row][i]);
             if(++begin_col > end_col) break;

             for(int i = begin_col; i <= end_col; ++i)//up to down
                  ans.push_back(matrix[i][end_row]);
             if(begin_row > --end_row) break;

             for(int i = end_row; i >= begin_row; --i)//right to left
                 ans.push_back(matrix[end_col][i]);
             if(begin_col > --end_col) break;

             for(int i = end_col; i >= begin_col; --i)//bottom to up
                 ans.push_back(matrix[i][begin_row]);
             if(++begin_row > end_row) break;
         }
         return ans;
     }
 };

 int main()
 {
     Solution s;
     vector<vector<int> > matrix;
     vector<int> v;
     for(int i = ; i <=; i++){
         v.push_back(i);
         if(i %  == ){
             matrix.push_back(v);
             v.clear();
         }
     }
     vector<int> ans = s.spiralOrder(matrix);
     for(int i = ; i < ans.size(); ++i)
         cout<< ans[i] <<endl;
     return ;
 }