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82 lines
2.2 KiB
Markdown
82 lines
2.2 KiB
Markdown
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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
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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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