Spiral Matrix
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return [1,2,3,6,9,8,7,4,5].
打印螺旋矩阵
逐个环的打印, 对于m *n的矩阵,环的个数是 (min(n,m)+1) / 2。对于每个环顺时针打印四条边。
注意的是：最后一个环可能只包含一行或者一列数据
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class
Solution {
public
:
vector<
int
> spiralOrder(vector<vector<
int
> > &matrix) {
int
m = matrix.size(), n;
if
(m != 0)n = matrix[0].size();
int
cycle = m > n ? (n+1)/2 : (m+1)/2;
//环的数目
vector<
int
>res;
int
a = n, b = m;
//a,b分别为当前环的宽度、高度
for
(
int
i = 0; i < cycle; i++, a -= 2, b -= 2)
{
//每个环的左上角起点是matrix[i][i],下面顺时针依次打印环的四条边
for
(
int
column = i; column < i+a; column++)
res.push_back(matrix[i][column]);
for
(
int
row = i+1; row < i+b; row++)
res.push_back(matrix[row][i+a-1]);
if
(a == 1 || b == 1)
break
;
//最后一个环只有一行或者一列
for
(
int
column = i+a-2; column >= i; column--)
res.push_back(matrix[i+b-1][column]);
for
(
int
row = i+b-2; row > i; row--)
res.push_back(matrix[row][i]);
}
return
res;
}
};
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Spiral Matrix II
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 ] ]
本质上和上一题是一样的，这里我们要用数字螺旋的去填充矩阵。同理，我们也是逐个环的填充，每个环顺时针逐条边填充 本文地址
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class
Solution {
public
:
vector<vector<
int
> > generateMatrix(
int
n) {
vector<vector<
int
> > matrix(n, vector<
int
>(n));
int
a = n;
//a为当前环的边长
int
val = 1;
for
(
int
i = 0; i < n/2; i++, a -= 2)
{
//每个环的左上角起点是matrix[i][i],下面顺时针依次填充环的四条边
for
(
int
column = i; column < i+a; column++)
matrix[i][column] = val++;
for
(
int
row = i+1; row < i+a; row++)
matrix[row][i+a-1] = val++;
for
(
int
column = i+a-2; column >= i; column--)
matrix[i+a-1][column] = val++;
for
(
int
row = i+a-2; row > i; row--)
matrix[row][i] = val++;
}
if
(n % 2)matrix[n/2][n/2] = val;
//n是奇数时，最后一个环只有一个数字
return
matrix;
}
};
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本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3774747.html，如需转载请自行联系原作者
