2021-03-18 02:05:13 +00:00
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2021-03-20 08:30:29 +00:00
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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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2021-03-20 07:48:03 +00:00
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### [59.螺旋矩阵 II](https://leetcode-cn.com/problems/spiral-matrix-ii)
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2021-03-18 02:05:13 +00:00
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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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*![螺旋矩阵](https://pic.leetcode-cn.com/1615813563-uUiWlF-file_1615813563382)*
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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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![](https://img-blog.csdnimg.cn/20210318095839543.gif)
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题目代码:
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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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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
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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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