leetcode 54. Spiral Matrix (矩阵蛇形取数)

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

思路：。。。再次让我回想起高一的暑假。。。。

/* ***********************************************
Author :111qqz
Created Time :2017年04月11日 星期二 19时42分05秒
File Name :54.cpp
************************************************ */
class Solution {
public:
    int n,m; //0右，1下，2左，3上
    int cal( int &x,int &y,int dir,vector<vector<bool> > & vis)
    {
	if (dir==0)
	{
	    if (y<=n-2&&!vis[x][y+1]) y++;
	    else
	    {
		dir++;
		x++;
	    }
	    return dir;
	}
	if (dir==1)
	{
	    if (x<=m-2&&!vis[x+1][y]) x++;
	    else
	    {
		dir++;
		y--;
	    }
	    return dir;
	}
	if (dir==2)
	{
	    if (y>=1&&!vis[x][y-1]) y--;
	    else
	    {
		dir++;
		x--;
	    }
	    return dir;
	}
	if (dir==3)
	{
	    if (x>=1&&!vis[x-1][y]) x--;
	    else
	    {
		dir = 0 ;
		y++;
	    }
	    return dir;
	}
    }
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
	vector<int>res;
	m = matrix.size();
	if (m==0) return res;
	n = matrix[0].size();
	if (n==0) return res;
	vector<vector<bool> >vis(m,vector<bool>(n,false));
	int x,y,dir;
	x = y = dir = 0 ;
	for ( int i = 0 ; i < n*m ;  i++)
	{
	    printf("x:%d y:%d\n",x,y);
	    res.push_back(matrix[x][y]);
	    vis[x][y] = true;
	    dir = cal(x,y,dir,vis);
	}
	return res;
    }
};