【leetcode】Spiral Matrix II （middle）

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 ]

]

 

思路：

我是直接套用了Spiral Matrix 的代码

class Solution {

public:

    vector<vector<int> > generateMatrix(int n) {

        vector<vector<int>> ans(n, vector<int>(n,0));



        int num = 1;

        for(int i = 0; i < (n + 1) / 2; i++)

        {

            //行 ----->

            for(int j = i; j < n - i; j++)

                ans[i][j] = num++;

            //列 向下

            for(int j = i + 1; j < n - i; j++)

                ans[j][n - 1 - i] = num++;

            //行 <-----------

            if(n - i - 1 <= i) //列号 一定要比向左时的列号小 防止重复

                break;

            for(int j = n - i - 2; j >= i; j--)

                ans[n - i - 1][j] = num++;

            //列 向上

            if(i >= n - 1 - i) //行号 一定要比向下时的行号大 防止重复

                break;

            for(int j = n - i - 2; j >= i + 1; j--)

                ans[j][i] = num++;

        }

        return ans;

    }

};

 

另一种写法的：

class Solution {

    public:

        vector<vector<int> > generateMatrix(int n) {

            vector<vector<int> > ret( n, vector<int>(n) );

            int k = 1, i = 0;

            while( k <= n * n )

            {

                int j = i;

                    // four steps

                while( j < n - i )             // 1. horizonal, left to right

                    ret[i][j++] = k++;

                j = i + 1;

                while( j < n - i )             // 2. vertical, top to bottom

                    ret[j++][n-i-1] = k++;

                j = n - i - 2;

                while( j > i )                  // 3. horizonal, right to left 

                    ret[n-i-1][j--] = k++;

                j = n - i - 1;

                while( j > i )                  // 4. vertical, bottom to  top 

                    ret[j--][i] = k++;

                i++;      // next loop

            }

            return ret;

        }

    };