Given an integer n, generate a square matrix filled with elements from 1 to n^2 in spiral order.
For example,
Given n = 3,
You should return the following matrix:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
与 54. Spiral Matrix 类似，这次给定一个整数n，以螺旋顺序填充元素到n^2的矩阵。
Java:
class Solution {
public int[][] generateMatrix(int n) {
int[][] res = new int[n][n];
int k = 1;
int top = 0, bottom = n - 1, left = 0, right = n - 1;
while (left < right && top < bottom) {
for (int j = left; j < right; j++) {
res[top][j] = k++;
}
for (int i = top; i < bottom; i++) {
res[i][right] = k++;
}
for (int j = right; j > left; j--) {
res[bottom][j] = k++;
}
for (int i = bottom; i > top; i--) {
res[i][left] = k++;
}
left++;
right--;
top++;
bottom--;
}
if (n % 2 != 0)
res[n / 2][n / 2] = k;
return res;
}
}
Python:
class Solution:
# @param matrix, a list of lists of integers
# @return a list of integers
def spiralOrder(self, matrix):
result = []
if matrix == []:
return result left, right, top, bottom = 0, len(matrix[0]) - 1, 0, len(matrix) - 1 while left <= right and top <= bottom:
for j in xrange(left, right + 1):
result.append(matrix[top][j])
for i in xrange(top + 1, bottom):
result.append(matrix[i][right])
for j in reversed(xrange(left, right + 1)):
if top < bottom:
result.append(matrix[bottom][j])
for i in reversed(xrange(top + 1, bottom)):
if left < right:
result.append(matrix[i][left])
left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1 return result
C++：
class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int> > res(n, vector<int>(n, 1));
int val = 1, p = n;
for (int i = 0; i < n / 2; ++i, p -= 2) {
for (int col = i; col < i + p; ++col)
res[i][col] = val++;
for (int row = i + 1; row < i + p; ++row)
res[row][i + p - 1] = val++;
for (int col = i + p - 2; col >= i; --col)
res[i + p - 1][col] = val++;
for (int row = i + p - 2; row > i; --row)
res[row][i] = val++;
}
if (n % 2 != 0) res[n / 2][n / 2] = val;
return res;
}
};
类似题目：
[LeetCode] 54. Spiral Matrix 螺旋矩阵