LeetCode Spiral Matrix II 带打印程序

Spiral Matrix II

Given an integern, generate a square matrix filled with elements from 1 ton2in spiral order.

For example,
Givenn=3,

You should return the following matrix:

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

有前面的Spiral Matrix作为基础就好做这道题了。

关键就是处理下标，要学会如何去跟踪下标，这个思维能力也需要锻炼。

锻炼好，那么就不需要20分钟就可以写出下面那么简练的程序了。

class Solution {
public:
	vector<vector<int> > generateMatrix(int n) {
		int i = 0, j = n-1;
		vector<vector<int> > res(n, vector<int>(n));

		int num = 1;
		for (; i < j; i++, j--)
		{
			for (int k = i; k <= j; k++)
			{
				res[i][k] = num;
				num++;
			}
			for (int k = i+1; k < j; k++)
			{
				res[k][j] = num;
				num++;
			}
			for (int k = j; k >= i; k--)
			{
				res[j][k] = num;
				num++;
			}
			for (int k = j-1; k > i; k--)
			{
				res[k][i] = num;
				num++;
			}
		}
		//n为基数，剩下一个空格的时候，不能重复填了。额外处理。
		if (i == j) res[i][j] = num;
		return res;
	}
};

测试程序：

int main() 
{
	Solution solu;
	vector<vector<int> > v = solu.generateMatrix(10);
	for (auto x:v)
	{
		for (auto y:x)
			cout<<y<<"\t";
		cout<<endl;
	}

	system("pause"); 
	return 0;
}

漂亮的打印结果：

//2014-1-29 update
	vector<vector<int> > generateMatrix(int n) 
	{
		vector<vector<int> > rs(n, vector<int>(n));
		int c = 1, i = 0;
		
		for (n--; i < n; i++, n--)
		{
			for (int j = i; j < n; j++, c++)
				rs[i][j] = c;
			for (int j = i; j < n; j++, c++)
				rs[j][n] = c;
			for (int j = n; j > i ; j--, c++)
				rs[n][j] = c;
			for (int j = n; j > i ; j--, c++)
				rs[j][i] = c;
		}
		if (i == n) rs[i][i] = c;
		return rs;
	}