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59. Spiral Matrix II
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‎markdown/59. Spiral Matrix II.md

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### [59\. Spiral Matrix II](https://leetcode.com/problems/spiral-matrix-ii/)
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Difficulty: **Medium**
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Given a positive integer _n_, generate a square matrix filled with elements from 1 to _n_<sup>2</sup> in spiral order.
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**Example:**
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```
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Input: 3
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Output:
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[
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[ 1, 2, 3 ],
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[ 8, 9, 4 ],
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[ 7, 6, 5 ]
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]
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```
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#### MySolution
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Language: **Java**
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```java
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class Solution {
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public int[][] generateMatrix(int n) {
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if (n == 0) {
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return new int[0][0];
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}
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int[][] matrix = new int[n][n];
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int x = 0; // 需要填充的坐标
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int y = 0; // 需要填充的坐标
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int direction = 0; // 0-右 1-下 2-左 3-上
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for (int i = 1; i <= n * n; i++) {
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matrix[x][y] = i; // 写
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// 判断下一个位置走向，索引定位到下一个位置
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switch (direction) {
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case 0:
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if (available(matrix, n, x, y + 1)) {
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y++;
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} else {
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direction = 1;
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x++;
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}
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break;
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case 1:
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if (available(matrix, n, x + 1, y)) {
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x++;
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} else {
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direction = 2;
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y--;
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}
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break;
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case 2:
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if (available(matrix, n, x, y - 1)) {
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y--;
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} else {
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direction = 3;
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x--;
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}
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break;
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case 3:
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if (available(matrix, n, x - 1, y)) {
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x--;
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} else {
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direction = 0;
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y++;
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}
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break;
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}
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}
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return matrix;
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}
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/**
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* 判断坐标是否可用
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*
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* @param n 大小
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* @param x 横坐标
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* @param y 纵坐标
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* @return 是否可用
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*/
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private boolean available(int[][] matrix, int n, int x, int y) {
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return !(x >= n || y >= n || x < 0 || y < 0 || matrix[x][y] > 0);
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}
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}
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```
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#### Recommend Solution
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Language: **Java**
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```java
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class Solution {
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public static int[][] generateMatrix(int n) {
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int[][] ret = new int[n][n];
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int left = 0, top = 0;
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int right = n - 1, down = n - 1;
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int count = 1;
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while (left <= right) {
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for (int j = left; j <= right; j++) {
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ret[top][j] = count++;
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}
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top++;
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for (int i = top; i <= down; i++) {
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ret[i][right] = count++;
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}
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right--;
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for (int j = right; j >= left; j--) {
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ret[down][j] = count++;
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}
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down--;
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for (int i = down; i >= top; i--) {
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ret[i][left] = count++;
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}
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left++;
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}
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return ret;
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}
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}
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```
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![](http://ww4.sinaimg.cn/large/006tNc79ly1g4yndj6hpfj31bk0py431.jpg)

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

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package leetcode._59_;
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/**
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* Created by zhangbo54 on 2019-03-04.
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*/
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public class Main {
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public static void main(String[] args) {
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Solution solution = new Solution();
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int[][] ints = solution.generateMatrix(0);
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for (int i = 0; i < ints.length; i++) {
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for (int j = 0; j < ints[i].length; j++) {
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System.out.print(ints[i][j] + ",");
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}
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System.out.println();
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}
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}
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}
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‎src/main/java/leetcode/_59_/Solution.java

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package leetcode._59_;
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class Solution {
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public int[][] generateMatrix(int n) {
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if (n == 0) {
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return new int[0][0];
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}
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int[][] matrix = new int[n][n];
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int x = 0; // 需要填充的坐标
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int y = 0; // 需要填充的坐标
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int direction = 0; // 0-右 1-下 2-左 3-上
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for (int i = 1; i <= n * n; i++) {
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matrix[x][y] = i; // 写
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// 判断下一个位置走向，索引定位到下一个位置
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switch (direction) {
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case 0:
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if (available(matrix, n, x, y + 1)) {
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y++;
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} else {
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direction = 1;
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x++;
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}
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break;
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case 1:
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if (available(matrix, n, x + 1, y)) {
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x++;
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} else {
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direction = 2;
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y--;
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}
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break;
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case 2:
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if (available(matrix, n, x, y - 1)) {
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y--;
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} else {
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direction = 3;
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x--;
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}
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break;
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case 3:
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if (available(matrix, n, x - 1, y)) {
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x--;
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} else {
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direction = 0;
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y++;
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}
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break;
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}
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}
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return matrix;
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}
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/**
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* 判断坐标是否可用
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*
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* @param n 大小
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* @param x 横坐标
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* @param y 纵坐标
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* @return 是否可用
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*/
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private boolean available(int[][] matrix, int n, int x, int y) {
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return !(x >= n || y >= n || x < 0 || y < 0 || matrix[x][y] > 0);
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}
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}

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

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### [59\. Spiral Matrix II](https://leetcode.com/problems/spiral-matrix-ii/)
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Difficulty: **Medium**
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Given a positive integer _n_, generate a square matrix filled with elements from 1 to _n_<sup>2</sup> in spiral order.
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8+
**Example:**
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10+
```
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Input: 3
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Output:
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[
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[ 1, 2, 3 ],
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[ 8, 9, 4 ],
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[ 7, 6, 5 ]
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]
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```
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#### MySolution
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Language: **Java**
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```java
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class Solution {
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public int[][] generateMatrix(int n) {
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if (n == 0) {
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return new int[0][0];
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}
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int[][] matrix = new int[n][n];
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int x = 0; // 需要填充的坐标
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int y = 0; // 需要填充的坐标
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int direction = 0; // 0-右 1-下 2-左 3-上
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for (int i = 1; i <= n * n; i++) {
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matrix[x][y] = i; // 写
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// 判断下一个位置走向，索引定位到下一个位置
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switch (direction) {
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case 0:
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if (available(matrix, n, x, y + 1)) {
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y++;
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} else {
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direction = 1;
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x++;
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}
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break;
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case 1:
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if (available(matrix, n, x + 1, y)) {
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x++;
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} else {
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direction = 2;
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y--;
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}
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break;
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case 2:
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if (available(matrix, n, x, y - 1)) {
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y--;
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} else {
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direction = 3;
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x--;
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}
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break;
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case 3:
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if (available(matrix, n, x - 1, y)) {
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x--;
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} else {
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direction = 0;
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y++;
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}
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break;
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}
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}
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return matrix;
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}
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/**
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* 判断坐标是否可用
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*
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* @param n 大小
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* @param x 横坐标
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* @param y 纵坐标
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* @return 是否可用
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*/
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private boolean available(int[][] matrix, int n, int x, int y) {
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return !(x >= n || y >= n || x < 0 || y < 0 || matrix[x][y] > 0);
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}
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}
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```
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#### Recommend Solution
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Language: **Java**
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```java
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class Solution {
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public static int[][] generateMatrix(int n) {
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int[][] ret = new int[n][n];
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int left = 0, top = 0;
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int right = n - 1, down = n - 1;
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int count = 1;
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while (left <= right) {
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for (int j = left; j <= right; j++) {
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ret[top][j] = count++;
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}
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top++;
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for (int i = top; i <= down; i++) {
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ret[i][right] = count++;
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}
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right--;
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for (int j = right; j >= left; j--) {
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ret[down][j] = count++;
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}
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down--;
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for (int i = down; i >= top; i--) {
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ret[i][left] = count++;
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}
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left++;
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}
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return ret;
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}
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}
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```
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![](http://ww4.sinaimg.cn/large/006tNc79ly1g4yndj6hpfj31bk0py431.jpg)

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