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algorithm-base/animation-simulation/数组篇/leetcode59螺旋矩阵2.md

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erfen
2021-03-18 02:05:13 +00:00
chefyuan
2021-03-20 08:30:29 +00:00
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>
> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
>
> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
chefyuan
2021-03-20 07:48:03 +00:00
### [59.螺旋矩阵 II](https://leetcode-cn.com/problems/spiral-matrix-ii)
erfen
2021-03-18 02:05:13 +00:00
给你一个正整数 `n` ，生成一个包含 `1` 到 `n2` 所有元素，且元素按顺时针顺序螺旋排列的 `n x n` 正方形矩阵 `matrix` 。
**示例 1：**
> 输入：n = 3
> 输出：[[1,2,3],[8,9,4],[7,6,5]]
**示例 2：**
> 输入：n = 1
> 输出：[[1]]
其实我们只要做过了螺旋矩阵 第一题，这个题目我们完全可以一下搞定，几乎没有进行更改，我们先来看下 **leetcode 54** 题的解析。
### leetcode 54 螺旋矩阵
题目描述
*给定一个包含 m* x n个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
示例一
> 输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
> 输出：[1,2,3,6,9,8,7,4,5]
示例二
> 输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
> 输出：[1,2,3,4,8,12,11,10,9,5,6,7]
这个题目很细非常细，思路很容易想到，但是要是完全实现也不是特别容易，我们一起分析下这个题目，我们可以这样理解，我们像剥洋葱似的一步步的剥掉外皮，直到遍历结束，见下图。
*![螺旋矩阵](https://pic.leetcode-cn.com/1615813563-uUiWlF-file_1615813563382)*
题目很容易理解，但是要想完全执行出来，也是不容易的，因为这里面的细节太多了，我们需要认真仔细的考虑边界。
我们也要考虑重复遍历的情况即什么时候跳出循环。刚才我们通过箭头知道了我们元素的遍历顺序，这个题目也就完成了一大半了，下面我们来讨论一下什么时候跳出循环，见下图。
注：这里需要注意的是，框框代表的是每个边界。
![](https://img-blog.csdnimg.cn/20210318095839543.gif)
题目代码：
添加了python代码（数组篇） 为数组篇文件夹下的代码增加了python语言版本
2021-07-10 04:20:02 +00:00
Java Code:
erfen
2021-03-18 02:05:13 +00:00
```java
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> arr = new ArrayList<>();
int left = 0, right = matrix[0].length-1;
int top = 0, down = matrix.length-1;
while (true) {
for (int i = left; i <= right; ++i) {
arr.add(matrix[top][i]);
}
top++;
if (top > down) break;
for (int i = top; i <= down; ++i) {
arr.add(matrix[i][right]);
}
right--;
if (left > right) break;
for (int i = right; i >= left; --i) {
arr.add(matrix[down][i]);
}
down--;
if (top > down) break;
for (int i = down; i >= top; --i) {
arr.add(matrix[i][left]);
}
left++;
if (left > right) break;
}
return arr;
}
}
```
添加了python代码（数组篇） 为数组篇文件夹下的代码增加了python语言版本
2021-07-10 04:20:02 +00:00
Python3 Code:
```python
from typing import List
class Solution:
def spiralOrder(self, matrix: List[List[int]])->List[int]:
arr = []
left = 0
right = len(matrix[0]) - 1
top = 0
down = len(matrix) - 1
while True:
for i in range(left, right + 1):
arr.append(matrix[top][i])
top += 1
if top > down:
break
for i in range(top, down + 1):
arr.append(matrix[i][right])
right -= 1
if left > right:
break
for i in range(right, left - 1, -1):
arr.append(matrix[down][i])
down -= 1
if top > down:
break
for i in range(down, top - 1, -1):
arr.append(matrix[i][left])
left += 1
if left > right:
break
return arr
```
erfen
2021-03-18 02:05:13 +00:00
数组篇增加部分C++代码
2021-07-15 15:14:17 +00:00
C++ Code:
```cpp
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector <int> arr;
int left = 0, right = matrix[0].size()-1;
int top = 0, down = matrix.size()-1;
while (true) {
for (int i = left; i <= right; ++i) {
arr.emplace_back(matrix[top][i]);
}
top++;
if (top > down) break;
for (int i = top; i <= down; ++i) {
arr.emplace_back(matrix[i][right]);
}
right--;
if (left > right) break;
for (int i = right; i >= left; --i) {
arr.emplace_back(matrix[down][i]);
}
down--;
if (top > down) break;
for (int i = down; i >= top; --i) {
arr.emplace_back(matrix[i][left]);
}
left++;
if (left > right) break;
}
return arr;
}
};
```
为数组篇 增加 Swift 实现
2021-07-17 04:13:15 +00:00
Swift Code:
```swift
class Solution {
func spiralOrder(_ matrix: [[Int]]) -> [Int] {
var arr:[Int] = []
var left = 0, right = matrix[0].count - 1
var top = 0, down = matrix.count - 1
while (true) {
for i in left...right {
arr.append(matrix[top][i])
}
top += 1
if top > down { break }
for i in top...down {
arr.append(matrix[i][right])
}
right -= 1
if left > right { break}
for i in stride(from: right, through: left, by: -1) {
arr.append(matrix[down][i])
}
down -= 1
if top > down { break}
for i in stride(from: down, through: top, by: -1) {
arr.append(matrix[i][left])
}
left += 1
if left > right { break}
}
return arr
}
}
```
erfen
2021-03-18 02:05:13 +00:00
我们仅仅是将 54 反过来了，往螺旋矩阵里面插值，下面我们直接看代码吧,大家可以也可以对其改进，大家可以思考一下，如果修改能够让代码更简洁！
添加了python代码（数组篇） 为数组篇文件夹下的代码增加了python语言版本
2021-07-10 04:20:02 +00:00
Java Code:
erfen
2021-03-18 02:05:13 +00:00
```java
class Solution {
public int[][] generateMatrix(int n) {
int[][] arr = new int[n][n];
int left = 0;
int right = n-1;
int top = 0;
int buttom = n-1;
int num = 1;
int numsize = n*n;
while (true) {
for (int i = left; i <= right; ++i) {
arr[top][i] = num++;
}
top++;
if (num > numsize) break;
for (int i = top; i <= buttom; ++i) {
arr[i][right] = num++;
}
right--;
if (num > numsize) break;
for (int i = right; i >= left; --i) {
arr[buttom][i] = num++;
}
buttom--;
if (num > numsize) break;
for (int i = buttom; i >= top; --i) {
arr[i][left] = num++;
}
left++;
if (num > numsize) break;
}
return arr;
}
}
```
添加了python代码（数组篇） 为数组篇文件夹下的代码增加了python语言版本
2021-07-10 04:20:02 +00:00
Python3 Code:
```python
from typing import List
import numpy as np
class Solution:
def generateMatrix(self, n: int)->List[List[int]]:
arr = np.array([[0] * n] * n)
left = 0
right = n - 1
top = 0
buttom = n - 1
num = 1
numsize = n * n
while True:
for i in range(left, right + 1):
arr[top][i] = num
num += 1
top += 1
if num > numsize:
break
for i in range(top, buttom + 1):
arr[i][right] = num
num += 1
right -= 1
if num > numsize:
break
for i in range(right, left - 1, -1):
arr[buttom][i] = num
num += 1
buttom -= 1
if num > numsize:
break
for i in range(buttom, top - 1, -1):
arr[i][left] = num
num += 1
left += 1
if num > numsize:
break
return arr.tolist()
```
数组篇增加部分C++代码
2021-07-15 15:14:17 +00:00
C++ Code:
```cpp
class Solution {
public:
vector<vector<int>> generateMatrix(int n) {
vector <vector <int>> arr(n, vector <int>(n));
int left = 0, right = n-1, top = 0, buttom = n - 1, num = 1, numsize = n * n;
while (true) {
for (int i = left; i <= right; ++i) {
arr[top][i] = num++;
}
top++;
if (num > numsize) break;
for (int i = top; i <= buttom; ++i) {
arr[i][right] = num++;
}
right--;
if (num > numsize) break;
for (int i = right; i >= left; --i) {
arr[buttom][i] = num++;
}
buttom--;
if (num > numsize) break;
for (int i = buttom; i >= top; --i) {
arr[i][left] = num++;
}
left++;
if (num > numsize) break;
}
return arr;
}
};
```
为数组篇 增加 Swift 实现
2021-07-17 04:13:15 +00:00
Swift Code:
```swift
class Solution {
func generateMatrix(_ n: Int) -> [[Int]] {
var arr:[[Int]] = Array.init(repeating: Array.init(repeating: 0, count: n), count: n)
var left = 0, right = n - 1
var top = 0, bottom = n - 1
var num = 1, numSize = n * n
while true {
for i in left...right {
arr[top][i] = num
num += 1
}
top += 1
if num > numSize { break}
for i in top...bottom {
arr[i][right] = num
num += 1
}
right -= 1
if num > numSize { break}
for i in stride(from: right, through: left, by: -1) {
arr[bottom][i] = num
num += 1
}
bottom -= 1
if num > numSize { break}
for i in stride(from: bottom, through: top, by: -1) {
arr[i][left] = num
num += 1
}
left += 1
if num > numSize { break}
}
return arr
}
}
```
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