LeetCode OJ 之 Spiral Matrix II （螺旋矩阵 - 二）

题目：

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 ]
]

给一个整数n，产生一个方阵，按螺旋顺序填入从1到 n^2这些数。

思路：

参考：http://blog.csdn.net/u012243115/article/details/42870849 。

代码：

class Solution {
public:
vector<vector<int> > generateMatrix(int n)
{
vector<vector<int> > result(n,vector<int>(n,0));
int rowBegin = 0 , rowEnd = n - 1;
int colBegin = 0 , colEnd = n - 1;
for(int num = 1 ; num <= n * n ; )
{
for(int j = colBegin ; j <= colEnd ; j++)
{
result[rowBegin][j] = num;
num++;
}
if(++rowBegin > rowEnd)
break;
for(int i = rowBegin ; i <= rowEnd ; i++ )
{
result[i][colEnd] = num;
num++;
}
if(--colEnd < colBegin)
break;
for(int j = colEnd ; j >= colEnd ; j--)
{
result[rowEnd][j] = num;
num++;
}
if(--rowEnd < rowBegin)
break;
for(int i = rowEnd ; i >= rowBegin ; i--)
{
result[i][colBegin] = num;
num++;
}
if(++colBegin > colEnd)
break;
}
return result;
}
};

由于是方阵，可以优化一下：

class Solution {
public:
vector<vector<int> > generateMatrix(int n)
{
vector<vector<int> > result(n,vector<int>(n,0));
int begin = 0 , end = n - 1;
int num = 1;
while(begin < end)
{
for(int j = begin ; j < end ; j++)
result[begin][j] = num++;
for(int i = begin ; i < end ; i++)
result[i][end] = num++;
for(int j = end ; j > begin ; j--)
result[end][j] = num++;
for(int i = end ; i > begin ; i--)
result[i][begin] = num++;
begin++;
end--;
}
if(begin == end)
result[begin][end] = num;
return result;
}
};