题目:
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:这道题和Spiral Matrix的模型和解法一样。要求我们将1~n^2的数字螺旋写入矩阵中。我们依然维护四个变量,rowBegin, rowEnd, colBegin, colEnd。然后去界定螺旋的边界。我们需要注意的是,如果n 为0,我们应该返回空的数组。并且在写入数字时,我们也需要判断目前的数字是否超过最大数字N^2,如果超过,说明我们不需要遍历了。还有最重要的一点,数组在未初始化时,是不允许使用下标操作符来改变数组元素的,所以我们一定要先初始化数组,这里初始化为N行N列,元素都为0的数组。
Attention:
1. 返回空数组 别忘了括号。
if(n == 0) return vector
>();
2. 初始化数组
vector
> ret(n, vector
(n, 0));
3. 要记得判断是否达到数字最大值。每次遍历都需要判断
if (num <= n^2)
{
//向右遍历添加
for (int j = colBegin; j <= colEnd; j++)
{
ret[rowBegin][j] = num;
num++;
}
}
复杂度:O(n^2) n为输入数字
AC Code:
class Solution {
public:
vector
> generateMatrix(int n) {
if(n == 0) return vector
>(); vector
> ret(n, vector
(n, 0)); int rowBegin = 0; int rowEnd = n - 1; int colBegin = 0; int colEnd = n - 1; int num = 1; while (rowBegin <= rowEnd && colBegin <= colEnd) { if (num <= n^2) { //向右遍历添加 for (int j = colBegin; j <= colEnd; j++) { ret[rowBegin][j] = num; num++; } } rowBegin++; if (num <= n^2) { //向下遍历添加 for (int i = rowBegin; i <= rowEnd; i++) { ret[i][colEnd] = num; num++; } } colEnd--; if (num <= n^2) { //向左遍历添加 for (int j = colEnd; j >= colBegin; j--) { ret[rowEnd][j] = num; num++; } } rowEnd--; if (num <= n^2) { //向上遍历添加 for (int i = rowEnd; i >= rowBegin; i--) { ret[i][colBegin] = num; num++; } } colBegin++; } return ret; } };