LeetCode:Spiral Matrix I II-白红宇

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

                            
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;
    
}
};
         
       

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

                            
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;
    
}
};
         
       

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