196 lines
5.8 KiB
Java
196 lines
5.8 KiB
Java
> 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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#### [54. 螺旋矩阵](https://leetcode-cn.com/problems/spiral-matrix/)
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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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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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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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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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