【leetcode】Spiral Matrix II (middle)
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 ]
]
思路:
我是直接套用了Spiral Matrix 的代码
class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int>> ans(n, vector<int>(n,)); int num = ;
for(int i = ; i < (n + ) / ; i++)
{
//行 ----->
for(int j = i; j < n - i; j++)
ans[i][j] = num++;
//列 向下
for(int j = i + ; j < n - i; j++)
ans[j][n - - i] = num++;
//行 <-----------
if(n - i - <= i) //列号 一定要比向左时的列号小 防止重复
break;
for(int j = n - i - ; j >= i; j--)
ans[n - i - ][j] = num++;
//列 向上
if(i >= n - - i) //行号 一定要比向下时的行号大 防止重复
break;
for(int j = n - i - ; j >= i + ; j--)
ans[j][i] = num++;
}
return ans;
}
};
另一种写法的:
class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int> > ret( n, vector<int>(n) );
int k = , i = ;
while( k <= n * n )
{
int j = i;
// four steps
while( j < n - i ) // 1. horizonal, left to right
ret[i][j++] = k++;
j = i + ;
while( j < n - i ) // 2. vertical, top to bottom
ret[j++][n-i-] = k++;
j = n - i - ;
while( j > i ) // 3. horizonal, right to left
ret[n-i-][j--] = k++;
j = n - i - ;
while( j > i ) // 4. vertical, bottom to top
ret[j--][i] = k++;
i++; // next loop
}
return ret;
}
};
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