【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 = 0, end_row = matrix[0].size() - 1;//begin_row is the row number, end_row is the remaind size of each row
        int begin_col = 0, end_col = matrix.size() - 1;//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 = 1; i <=9; i++){
        v.push_back(i);
        if(i % 3 == 0){
            matrix.push_back(v);
            v.clear();
        }
    }
    vector<int> ans = s.spiralOrder(matrix);
    for(int i = 0; i < ans.size(); ++i)
        cout<< ans[i] <<endl;
    return 0;        
}