Spiral Matrix I&&II

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].

这道题挺简单的，基本上算是一次性写出来的，就是设立一个对应的标志数组，然后按照螺旋的规则来遍历。

代码如下：

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> res;
        int height=matrix.size();
        if( height==0)
            return res;
        int width=matrix[0].size();
        vector<vector<int>> flag(height,vector<int>(width,1));//用来记录是否走过
        int m=0;
        int n=0;
        flag[0][0]=0;
        res.push_back(matrix[0][0]);
        int step=1;
        while(step!= height* width)
        {
            while(n+1<width&&flag[m][n+1])
            {
                flag[m][n+1]=0;
                res.push_back(matrix[m][n+1]);
                n+=1;
                step++;
            }
            while(m+1<height&&flag[m+1][n])
            {
                flag[m+1][n]=0;
                res.push_back(matrix[m+1][n]);
                m+=1;
                step++;
            }
            while(n-1>=0&&flag[m][n-1])
            {
                flag[m][n-1]=0;
                res.push_back(matrix[m][n-1]);
                n-=1;
                step++;
            }
            while(m-1>=0&&flag[m-1][n])
            {
                flag[m-1][n]=0;
                res.push_back(matrix[m-1][n]);
                m-=1;
                step++;
            }
        }
        return res;
    }
};

Given an integer n, generate a square matrix filled with elements from 1 to n² in spiral order.

For example,
Given n = 3,

You should return the following matrix:

[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

　I写出来了的话，II就更简单了，在I上改一下就行了，代码如下：

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> flag(n,vector<int>(n,0));//用来记录是否走过
        if(n==0)
            return flag;
        int height=0;
        int width=0;
        int step=1;
        flag[0][0]=1;
        while(step!= n*n)
        {
            while(width+1<n&&flag[height][width+1]==0)
            {
                width+=1;
                step++;
                flag[height][width]=step;
            }
            while(height+1<n&&flag[height+1][width]==0)
            {
                height+=1;
                step++;
                flag[height][width]=step;
            }
            while(width-1>=0&&flag[height][width-1]==0)
            {
                width-=1;
                step++;
                flag[height][width]=step;
            }
            while(height-1>=0&&flag[height-1][width]==0)
            {
                height-=1;
                step++;
                flag[height][width]=step;
            }
        }
        return flag;
        
    }
};