学步园

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].

思路：打印如螺旋桨，每个矩形框的起始点为(x,x)，如果row>2*x||col>2*x则在这个矩形框打印结束。每个矩形框的打印路径为(x,x)---->(col-x-1,x)--->(col-x-1,row-x-1)---->(x,row-x-1)----->(x,x)

class Solution {
public:
    void spiralOrderHelper(vector<int> &res, int row, int col, int startX, int startY,vector<vector<int> > matrix) {
        int i,j,p,q;
        for(i=startX; i<col-startX; ++i) { //从左到右打印
            res.push_back(matrix[startY][i]);
        }
        for(j=startY+1; j<row-startY; ++j) {     //从上到下打印
            res.push_back(matrix[j][col-startX-1]);
        }
        for(p=col-startX-2; j-1 > startY && p>=startX; p--) { //从右到左打印
            res.push_back(matrix[row-startY-1][p]);
        }
        for(q=row-startY-2; startX < col-startX-1 && q>startY; q--) { //从下到上打印
            res.push_back(matrix[q][startX]);
        }
    }
    vector<int> spiralOrder(vector<vector<int> > &matrix) {
        vector<int> res;
        res.clear();
        if (matrix.empty()) {
            return res;
        }
        int row = matrix.size();
        int col = matrix[0].size();
        int circle = 0;
        int startX = circle,startY = circle;
        while(row > 2*startX && col > 2*startY) {
            spiralOrderHelper(res,row,col,startX,startY,matrix);
            circle++;
            startX = circle, startY = circle;
        }
    }
};