Skip to content

Commit caedf40

Browse files
coderbeancoderbean
committed
59. Spiral Matrix II
1 parent f6d30f2 commit caedf40

File tree

4 files changed

+328
-0
lines changed

4 files changed

+328
-0
lines changed

‎markdown/59. Spiral Matrix II.md

Lines changed: 123 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,123 @@
1+
### [59\. Spiral Matrix II](https://leetcode.com/problems/spiral-matrix-ii/)
2+
3+
Difficulty: **Medium**
4+
5+
6+
Given a positive integer _n_, generate a square matrix filled with elements from 1 to _n_<sup>2</sup> in spiral order.
7+
8+
**Example:**
9+
10+
```
11+
Input: 3
12+
Output:
13+
[
14+
[ 1, 2, 3 ],
15+
[ 8, 9, 4 ],
16+
[ 7, 6, 5 ]
17+
]
18+
```
19+
20+
21+
#### MySolution
22+
23+
Language: **Java**
24+
25+
```java
26+
class Solution {
27+
public int[][] generateMatrix(int n) {
28+
if (n == 0) {
29+
return new int[0][0];
30+
}
31+
int[][] matrix = new int[n][n];
32+
int x = 0; // 需要填充的坐标
33+
int y = 0; // 需要填充的坐标
34+
int direction = 0; // 0-右 1-下 2-左 3-上
35+
for (int i = 1; i <= n * n; i++) {
36+
matrix[x][y] = i; // 写
37+
// 判断下一个位置走向，索引定位到下一个位置
38+
switch (direction) {
39+
case 0:
40+
if (available(matrix, n, x, y + 1)) {
41+
y++;
42+
} else {
43+
direction = 1;
44+
x++;
45+
}
46+
break;
47+
case 1:
48+
if (available(matrix, n, x + 1, y)) {
49+
x++;
50+
} else {
51+
direction = 2;
52+
y--;
53+
}
54+
break;
55+
case 2:
56+
if (available(matrix, n, x, y - 1)) {
57+
y--;
58+
} else {
59+
direction = 3;
60+
x--;
61+
}
62+
break;
63+
case 3:
64+
if (available(matrix, n, x - 1, y)) {
65+
x--;
66+
} else {
67+
direction = 0;
68+
y++;
69+
}
70+
break;
71+
}
72+
}
73+
return matrix;
74+
}
75+
76+
/**
77+
* 判断坐标是否可用
78+
*
79+
* @param n 大小
80+
* @param x 横坐标
81+
* @param y 纵坐标
82+
* @return 是否可用
83+
*/
84+
private boolean available(int[][] matrix, int n, int x, int y) {
85+
return !(x >= n || y >= n || x < 0 || y < 0 || matrix[x][y] > 0);
86+
}
87+
}
88+
```
89+
90+
#### Recommend Solution
91+
92+
Language: **Java**
93+
94+
```java
95+
class Solution {
96+
public static int[][] generateMatrix(int n) {
97+
int[][] ret = new int[n][n];
98+
int left = 0, top = 0;
99+
int right = n - 1, down = n - 1;
100+
int count = 1;
101+
while (left <= right) {
102+
for (int j = left; j <= right; j++) {
103+
ret[top][j] = count++;
104+
}
105+
top++;
106+
for (int i = top; i <= down; i++) {
107+
ret[i][right] = count++;
108+
}
109+
right--;
110+
for (int j = right; j >= left; j--) {
111+
ret[down][j] = count++;
112+
}
113+
down--;
114+
for (int i = down; i >= top; i--) {
115+
ret[i][left] = count++;
116+
}
117+
left++;
118+
}
119+
return ret;
120+
}
121+
}
122+
```
123+
![](http://ww4.sinaimg.cn/large/006tNc79ly1g4yndj6hpfj31bk0py431.jpg)

‎src/main/java/leetcode/_59_/Main.java

Lines changed: 18 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
1+
package leetcode._59_;
2+
3+
/**
4+
* Created by zhangbo54 on 2019-03-04.
5+
*/
6+
public class Main {
7+
public static void main(String[] args) {
8+
Solution solution = new Solution();
9+
int[][] ints = solution.generateMatrix(0);
10+
for (int i = 0; i < ints.length; i++) {
11+
for (int j = 0; j < ints[i].length; j++) {
12+
System.out.print(ints[i][j] + ",");
13+
}
14+
System.out.println();
15+
}
16+
}
17+
}
18+

‎src/main/java/leetcode/_59_/Solution.java

Lines changed: 64 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,64 @@
1+
package leetcode._59_;
2+
3+
class Solution {
4+
public int[][] generateMatrix(int n) {
5+
if (n == 0) {
6+
return new int[0][0];
7+
}
8+
int[][] matrix = new int[n][n];
9+
int x = 0; // 需要填充的坐标
10+
int y = 0; // 需要填充的坐标
11+
int direction = 0; // 0-右 1-下 2-左 3-上
12+
for (int i = 1; i <= n * n; i++) {
13+
matrix[x][y] = i; // 写
14+
// 判断下一个位置走向，索引定位到下一个位置
15+
switch (direction) {
16+
case 0:
17+
if (available(matrix, n, x, y + 1)) {
18+
y++;
19+
} else {
20+
direction = 1;
21+
x++;
22+
}
23+
break;
24+
case 1:
25+
if (available(matrix, n, x + 1, y)) {
26+
x++;
27+
} else {
28+
direction = 2;
29+
y--;
30+
}
31+
break;
32+
case 2:
33+
if (available(matrix, n, x, y - 1)) {
34+
y--;
35+
} else {
36+
direction = 3;
37+
x--;
38+
}
39+
break;
40+
case 3:
41+
if (available(matrix, n, x - 1, y)) {
42+
x--;
43+
} else {
44+
direction = 0;
45+
y++;
46+
}
47+
break;
48+
}
49+
}
50+
return matrix;
51+
}
52+
53+
/**
54+
* 判断坐标是否可用
55+
*
56+
* @param n 大小
57+
* @param x 横坐标
58+
* @param y 纵坐标
59+
* @return 是否可用
60+
*/
61+
private boolean available(int[][] matrix, int n, int x, int y) {
62+
return !(x >= n || y >= n || x < 0 || y < 0 || matrix[x][y] > 0);
63+
}
64+
}

‎src/main/java/leetcode/_59_/solution.md

Lines changed: 123 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,123 @@
1+
### [59\. Spiral Matrix II](https://leetcode.com/problems/spiral-matrix-ii/)
2+
3+
Difficulty: **Medium**
4+
5+
6+
Given a positive integer _n_, generate a square matrix filled with elements from 1 to _n_<sup>2</sup> in spiral order.
7+
8+
**Example:**
9+
10+
```
11+
Input: 3
12+
Output:
13+
[
14+
[ 1, 2, 3 ],
15+
[ 8, 9, 4 ],
16+
[ 7, 6, 5 ]
17+
]
18+
```
19+
20+
21+
#### MySolution
22+
23+
Language: **Java**
24+
25+
```java
26+
class Solution {
27+
public int[][] generateMatrix(int n) {
28+
if (n == 0) {
29+
return new int[0][0];
30+
}
31+
int[][] matrix = new int[n][n];
32+
int x = 0; // 需要填充的坐标
33+
int y = 0; // 需要填充的坐标
34+
int direction = 0; // 0-右 1-下 2-左 3-上
35+
for (int i = 1; i <= n * n; i++) {
36+
matrix[x][y] = i; // 写
37+
// 判断下一个位置走向，索引定位到下一个位置
38+
switch (direction) {
39+
case 0:
40+
if (available(matrix, n, x, y + 1)) {
41+
y++;
42+
} else {
43+
direction = 1;
44+
x++;
45+
}
46+
break;
47+
case 1:
48+
if (available(matrix, n, x + 1, y)) {
49+
x++;
50+
} else {
51+
direction = 2;
52+
y--;
53+
}
54+
break;
55+
case 2:
56+
if (available(matrix, n, x, y - 1)) {
57+
y--;
58+
} else {
59+
direction = 3;
60+
x--;
61+
}
62+
break;
63+
case 3:
64+
if (available(matrix, n, x - 1, y)) {
65+
x--;
66+
} else {
67+
direction = 0;
68+
y++;
69+
}
70+
break;
71+
}
72+
}
73+
return matrix;
74+
}
75+
76+
/**
77+
* 判断坐标是否可用
78+
*
79+
* @param n 大小
80+
* @param x 横坐标
81+
* @param y 纵坐标
82+
* @return 是否可用
83+
*/
84+
private boolean available(int[][] matrix, int n, int x, int y) {
85+
return !(x >= n || y >= n || x < 0 || y < 0 || matrix[x][y] > 0);
86+
}
87+
}
88+
```
89+
90+
#### Recommend Solution
91+
92+
Language: **Java**
93+
94+
```java
95+
class Solution {
96+
public static int[][] generateMatrix(int n) {
97+
int[][] ret = new int[n][n];
98+
int left = 0, top = 0;
99+
int right = n - 1, down = n - 1;
100+
int count = 1;
101+
while (left <= right) {
102+
for (int j = left; j <= right; j++) {
103+
ret[top][j] = count++;
104+
}
105+
top++;
106+
for (int i = top; i <= down; i++) {
107+
ret[i][right] = count++;
108+
}
109+
right--;
110+
for (int j = right; j >= left; j--) {
111+
ret[down][j] = count++;
112+
}
113+
down--;
114+
for (int i = down; i >= top; i--) {
115+
ret[i][left] = count++;
116+
}
117+
left++;
118+
}
119+
return ret;
120+
}
121+
}
122+
```
123+
![](http://ww4.sinaimg.cn/large/006tNc79ly1g4yndj6hpfj31bk0py431.jpg)

0 commit comments

Comments
 (0)
pFad - Phonifier reborn

Pfad - The Proxy pFad of © 2024 Garber Painting. All rights reserved.

Note: This service is not intended for secure transactions such as banking, social media, email, or purchasing. Use at your own risk. We assume no liability whatsoever for broken pages.


Alternative Proxies:

Alternative Proxy

pFad Proxy

pFad v3 Proxy

pFad v4 Proxy