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# Conflicts: # animation-simulation/数组篇/leetcode219数组中重复元素2.md # animation-simulation/数组篇/leetcode59螺旋矩阵2.md # animation-simulation/数组篇/leetcode66加一.md # animation-simulation/数组篇/leetcode75颜色分类.md
379 lines
11 KiB
Java
379 lines
11 KiB
Java
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> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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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。
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>
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> 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。
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### [59.螺旋矩阵 II](https://leetcode-cn.com/problems/spiral-matrix-ii)
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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` 。
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**示例 1：**
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> 输入：n = 3
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> 输出：[[1,2,3],[8,9,4],[7,6,5]]
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**示例 2：**
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> 输入：n = 1
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> 输出：[[1]]
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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。
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### leetcode 54 螺旋矩阵
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题目描述
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*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。
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示例一
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> 输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
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> 输出：[1,2,3,6,9,8,7,4,5]
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示例二
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> 输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
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> 输出：[1,2,3,4,8,12,11,10,9,5,6,7]
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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。
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**
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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。
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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。
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注：这里需要注意的是，框框代表的是每个边界。
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题目代码：
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Java Code:
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```java
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class Solution {
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public List<Integer> spiralOrder(int[][] matrix) {
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List<Integer> arr = new ArrayList<>();
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int left = 0, right = matrix[0].length-1;
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int top = 0, down = matrix.length-1;
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while (true) {
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for (int i = left; i <= right; ++i) {
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arr.add(matrix[top][i]);
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}
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top++;
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if (top > down) break;
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for (int i = top; i <= down; ++i) {
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arr.add(matrix[i][right]);
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}
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right--;
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if (left > right) break;
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for (int i = right; i >= left; --i) {
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arr.add(matrix[down][i]);
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}
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down--;
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if (top > down) break;
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for (int i = down; i >= top; --i) {
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arr.add(matrix[i][left]);
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}
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left++;
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if (left > right) break;
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}
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return arr;
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}
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}
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```
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Python3 Code:
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```python
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from typing import List
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class Solution:
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def spiralOrder(self, matrix: List[List[int]])->List[int]:
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arr = []
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left = 0
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right = len(matrix[0]) - 1
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top = 0
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down = len(matrix) - 1
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while True:
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for i in range(left, right + 1):
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arr.append(matrix[top][i])
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top += 1
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if top > down:
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break
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for i in range(top, down + 1):
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arr.append(matrix[i][right])
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right -= 1
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if left > right:
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break
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for i in range(right, left - 1, -1):
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arr.append(matrix[down][i])
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down -= 1
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if top > down:
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break
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for i in range(down, top - 1, -1):
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arr.append(matrix[i][left])
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left += 1
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if left > right:
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break
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return arr
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```
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C++ Code:
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```cpp
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class Solution {
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public:
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vector<int> spiralOrder(vector<vector<int>>& matrix) {
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vector <int> arr;
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int left = 0, right = matrix[0].size()-1;
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int top = 0, down = matrix.size()-1;
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while (true) {
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for (int i = left; i <= right; ++i) {
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arr.emplace_back(matrix[top][i]);
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}
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top++;
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if (top > down) break;
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for (int i = top; i <= down; ++i) {
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arr.emplace_back(matrix[i][right]);
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}
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right--;
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if (left > right) break;
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for (int i = right; i >= left; --i) {
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arr.emplace_back(matrix[down][i]);
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}
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down--;
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if (top > down) break;
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for (int i = down; i >= top; --i) {
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arr.emplace_back(matrix[i][left]);
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}
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left++;
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if (left > right) break;
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}
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return arr;
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}
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};
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```
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Swift Code:
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```swift
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class Solution {
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func spiralOrder(_ matrix: [[Int]]) -> [Int] {
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var arr:[Int] = []
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var left = 0, right = matrix[0].count - 1
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var top = 0, down = matrix.count - 1
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while (true) {
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for i in left...right {
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arr.append(matrix[top][i])
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}
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top += 1
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if top > down { break }
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for i in top...down {
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arr.append(matrix[i][right])
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}
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right -= 1
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if left > right { break}
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for i in stride(from: right, through: left, by: -1) {
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arr.append(matrix[down][i])
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}
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down -= 1
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if top > down { break}
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for i in stride(from: down, through: top, by: -1) {
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arr.append(matrix[i][left])
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}
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left += 1
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if left > right { break}
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}
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return arr
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}
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}
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```
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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！
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Java Code:
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```java
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class Solution {
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public int[][] generateMatrix(int n) {
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int[][] arr = new int[n][n];
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int left = 0;
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int right = n-1;
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int top = 0;
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int buttom = n-1;
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int num = 1;
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int numsize = n*n;
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while (true) {
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for (int i = left; i <= right; ++i) {
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arr[top][i] = num++;
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}
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top++;
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if (num > numsize) break;
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for (int i = top; i <= buttom; ++i) {
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arr[i][right] = num++;
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}
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right--;
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if (num > numsize) break;
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for (int i = right; i >= left; --i) {
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arr[buttom][i] = num++;
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}
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buttom--;
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if (num > numsize) break;
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for (int i = buttom; i >= top; --i) {
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arr[i][left] = num++;
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}
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left++;
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if (num > numsize) break;
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}
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return arr;
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}
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}
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```
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Python3 Code:
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```python
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from typing import List
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import numpy as np
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class Solution:
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def generateMatrix(self, n: int)->List[List[int]]:
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arr = np.array([[0] * n] * n)
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left = 0
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right = n - 1
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top = 0
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buttom = n - 1
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num = 1
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numsize = n * n
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while True:
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for i in range(left, right + 1):
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arr[top][i] = num
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num += 1
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top += 1
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if num > numsize:
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break
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for i in range(top, buttom + 1):
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arr[i][right] = num
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num += 1
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right -= 1
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if num > numsize:
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break
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for i in range(right, left - 1, -1):
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arr[buttom][i] = num
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num += 1
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buttom -= 1
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if num > numsize:
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break
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for i in range(buttom, top - 1, -1):
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arr[i][left] = num
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num += 1
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left += 1
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if num > numsize:
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break
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return arr.tolist()
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```
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C++ Code:
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```cpp
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class Solution {
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public:
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vector<vector<int>> generateMatrix(int n) {
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vector <vector <int>> arr(n, vector <int>(n));
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int left = 0, right = n-1, top = 0, buttom = n - 1, num = 1, numsize = n * n;
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while (true) {
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for (int i = left; i <= right; ++i) {
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arr[top][i] = num++;
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}
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top++;
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if (num > numsize) break;
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for (int i = top; i <= buttom; ++i) {
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arr[i][right] = num++;
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}
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right--;
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if (num > numsize) break;
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for (int i = right; i >= left; --i) {
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arr[buttom][i] = num++;
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}
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buttom--;
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if (num > numsize) break;
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for (int i = buttom; i >= top; --i) {
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arr[i][left] = num++;
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}
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left++;
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if (num > numsize) break;
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}
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return arr;
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}
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};
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```
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Swift Code:
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```swift
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class Solution {
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func generateMatrix(_ n: Int) -> [[Int]] {
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var arr:[[Int]] = Array.init(repeating: Array.init(repeating: 0, count: n), count: n)
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var left = 0, right = n - 1
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var top = 0, bottom = n - 1
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var num = 1, numSize = n * n
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while true {
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for i in left...right {
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arr[top][i] = num
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num += 1
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}
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top += 1
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if num > numSize { break}
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for i in top...bottom {
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arr[i][right] = num
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num += 1
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}
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right -= 1
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if num > numSize { break}
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for i in stride(from: right, through: left, by: -1) {
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arr[bottom][i] = num
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num += 1
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}
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bottom -= 1
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if num > numSize { break}
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for i in stride(from: bottom, through: top, by: -1) {
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arr[i][left] = num
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num += 1
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}
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left += 1
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if num > numSize { break}
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}
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return arr
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}
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}
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```
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