## Spiral Matrix II

### 描述

Given an integer n, generate a square matrix filled with elements from 1 to n^2 in spiral order.

For example, Given n = 3,

You should return the following matrix:

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


### 代码1

// Spiral Matrix II
// 时间复杂度O(n^2)，空间复杂度O(n^2)
public class Solution {
public int[][] generateMatrix(int n) {
int[][] matrix = new int[n][n];
int begin = 0, end = n - 1;
int num = 1;

while (begin < end) {
for (int j = begin; j < end; ++j) matrix[begin][j] = num++;
for (int i = begin; i < end; ++i) matrix[i][end] = num++;
for (int j = end; j > begin; --j) matrix[end][j] = num++;
for (int i = end; i > begin; --i) matrix[i][begin] = num++;
++begin;
--end;
}

if (begin == end) matrix[begin][begin] = num;

return matrix;
}
}

### 代码2

// LeetCode, Spiral Matrix II
// @author 龚陆安 (http://weibo.com/luangong)
// 时间复杂度O(n^2)，空间复杂度O(n^2)
class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector< vector<int> > matrix(n, vector<int>(n));
if (n == 0) return matrix;
int beginX = 0, endX = n - 1;
int beginY = 0, endY = n - 1;
int num = 1;
while (true) {
for (int j = beginX; j <= endX; ++j) matrix[beginY][j] = num++;
if (++beginY > endY) break;

for (int i = beginY; i <= endY; ++i) matrix[i][endX] = num++;
if (beginX > --endX) break;

for (int j = endX; j >= beginX; --j) matrix[endY][j] = num++;
if (beginY > --endY) break;

for (int i = endY; i >= beginY; --i) matrix[i][beginX] = num++;
if (++beginX > endX) break;
}
return matrix;
}
};