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