Difficulty: Medium
Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
Example:
Input: 3
Output:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
Language: Java
class Solution {
public int[][] generateMatrix(int n) {
if (n == 0) {
return new int[0][0];
}
int[][] matrix = new int[n][n];
int x = 0; // 需要填充的坐标
int y = 0; // 需要填充的坐标
int direction = 0; // 0-右 1-下 2-左 3-上
for (int i = 1; i <= n * n; i++) {
matrix[x][y] = i; // 写
// 判断下一个位置走向，索引定位到下一个位置
switch (direction) {
case 0:
if (available(matrix, n, x, y + 1)) {
y++;
} else {
direction = 1;
x++;
}
break;
case 1:
if (available(matrix, n, x + 1, y)) {
x++;
} else {
direction = 2;
y--;
}
break;
case 2:
if (available(matrix, n, x, y - 1)) {
y--;
} else {
direction = 3;
x--;
}
break;
case 3:
if (available(matrix, n, x - 1, y)) {
x--;
} else {
direction = 0;
y++;
}
break;
}
}
return matrix;
}
/**
* 判断坐标是否可用
*
* @param n 大小
* @param x 横坐标
* @param y 纵坐标
* @return 是否可用
*/
private boolean available(int[][] matrix, int n, int x, int y) {
return !(x >= n || y >= n || x < 0 || y < 0 || matrix[x][y] > 0);
}
}Language: Java
class Solution {
public static int[][] generateMatrix(int n) {
int[][] ret = new int[n][n];
int left = 0, top = 0;
int right = n - 1, down = n - 1;
int count = 1;
while (left <= right) {
for (int j = left; j <= right; j++) {
ret[top][j] = count++;
}
top++;
for (int i = top; i <= down; i++) {
ret[i][right] = count++;
}
right--;
for (int j = right; j >= left; j--) {
ret[down][j] = count++;
}
down--;
for (int i = down; i >= top; i--) {
ret[i][left] = count++;
}
left++;
}
return ret;
}
}