mirror of
https://github.com/chefyuan/algorithm-base.git
synced 2024-12-01 00:08:53 +00:00
82 lines
2.2 KiB
Markdown
82 lines
2.2 KiB
Markdown
|
### 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。
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
**
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
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。
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
题目代码：
|
|||
|
|
|
|||
|
|
```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;
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
|
|||
|
|
|