leetcode 59. Spiral Matrix II 螺旋方式写矩阵

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 ]
]

和上一道题leetcode 54. Spiral Matrix 螺旋方式读取矩阵相反的是，本题要求的是按照螺旋的方式写矩阵，实现方法一模一样。

代码如下：

public class Solution
{
    public int[][] generateMatrix(int n) 
    {
        if(n==0)
            return new int[0][0];

        int [][]matrix=new int[n][n];
        int count=1;
        int top=0,bottom=matrix.length-1;
        int left=0,right=matrix[0].length-1;
        while(true)
        {
            for(int i=left;i<=right;i++)
                matrix[top][i]=count++;
            top++;
            if(top>bottom)
                break;

            for(int i=top;i<=bottom;i++)
                matrix[i][right]=count++;
            right--;
            if(left>right)
                break;

            for(int i=right;i>=left;i--)
                matrix[bottom][i]=count++;
            bottom--;
            if(top>bottom)
                break;

            for(int i=bottom;i>=top;i--)
                matrix[i][left]=count++;
            left++;
            if(left>right)
                break;  
        }   
        return matrix;
    }
}

下面是C++的做法，这个和上一道题螺旋方式读取矩阵的做法一模一样

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;


class Solution
{
public:
    vector<vector<int>> generateMatrix(int n)
    {
        vector<vector<int>> mat(n,vector<int>(n,0));
        if (n <= 0)
            return mat;

        int count = 1 , left = 0, right = mat[0].size() - 1;
        int up = 0, down = mat.size() - 1;
        while (true)
        {
            for (int i = left; i <= right; i++)
                mat[up][i] = count++;
            up++;
            if (up > down)
                break;

            for (int i = up; i <= down; i++)
                mat[i][right] = count++;
            right--;
            if (left > right)
                break;

            for (int i = right; i >= left; i--)
                mat[down][i] = count++;
            down--;
            if (up > down)
                break;

            for (int i = down; i >= up; i--)
                mat[i][left] = count++;
            left++;
            if (left > right)
                break;
        }
        return mat;
    }
};