二刷54

Author: diguage

docs/0054-spiral-matrix.adoc

@@ -30,10 +30,36 @@ Given a matrix of _m_ x _n_ elements (_m_ rows, _n_ columns), return all element
 *Output:* [1,2,3,4,8,12,11,10,9,5,6,7]
 ----
 
+== 思路分析
+
+从回溯思想得到启发，使用递归来逐层推进。每次方法调用只负责指定层的遍历，向里推进层次的工作，交给递归来完成。这样避免了复杂的判断。
+
+image::images/0054-01.png[{image_attr}]
 
 [[src-0054]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0054_SpiralMatrix.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0054_SpiralMatrix_2.java[tag=answer]
+----
+--
+====
+
+== 参考资料
+
+. https://leetcode.cn/problems/spiral-matrix/solutions/275393/luo-xuan-ju-zhen-by-leetcode-solution/?envType=study-plan-v2&envId=selected-coding-interview[54. 螺旋矩阵 - 官方题解^]
+. https://leetcode.cn/problems/spiral-matrix/solutions/2362055/54-luo-xuan-ju-zhen-mo-ni-qing-xi-tu-jie-juvi/?envType=study-plan-v2&envId=selected-coding-interview[54. 螺旋矩阵 - 模拟，清晰图解^]
 

docs/images/0054-01.png

@@ -555,6 +555,11 @@
 |{doc_base_url}/0142-linked-list-cycle-ii.adoc[题解]
 |快慢指针
 
+|{counter:codes}
+|{leetcode_base_url}/spiral-matrix/[54. Spiral Matrix^]
+|{doc_base_url}/0054-spiral-matrix.adoc[题解]
+|从回溯得到启发，使用递归来完成层次的遍历。
+
 |===
 
 截止目前，本轮练习一共完成 {codes} 道题。

logbook/202406.adoc

@@ -42,6 +42,9 @@ public class _0054_SpiralMatrix {
      * Runtime: 0 ms, faster than 100.00% of Java online submissions for Spiral Matrix.
      *
      * Memory Usage: 34.4 MB, less than 100.00% of Java online submissions for Spiral Matrix.
+     *
+     * @author D瓜哥 · https://www.diguage.com
+     * @since 2019-10-26 00:51:20
      */
     public List<Integer> spiralOrder(int[][] matrix) {
         if (Objects.isNull(matrix) || matrix.length == 0) {

src/main/java/com/diguage/algo/leetcode/_0054_SpiralMatrix.java

@@ -0,0 +1,47 @@
+package com.diguage.algo.leetcode;
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class _0054_SpiralMatrix_2 {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-09-14 17:42:39
+   */
+  public List<Integer> spiralOrder(int[][] matrix) {
+    int row = matrix.length;
+    int column = matrix[0].length;
+    List<Integer> result = new ArrayList<>(row * column);
+    bfs(matrix, result, 0, 0, row, column);
+    return result;
+  }
+
+  private void bfs(int[][] matrix, List<Integer> result,
+                   int row, int column,
+                   int rLen, int cLen) {
+    if (rLen <= 0 || cLen <= 0) {
+      return;
+    }
+    for (int i = column; i < column + cLen; i++) {
+      result.add(matrix[row][i]);
+    }
+    for (int i = row + 1; i < row + rLen; i++) {
+      result.add(matrix[i][column + cLen - 1]);
+    }
+    // 不想增加复杂判断了，数量足够就直接返回
+    // 如果不返回，在最后只剩下一层且有多个元素时，中间元素会被重复添加
+    if (result.size() == matrix.length * matrix[0].length) {
+      return;
+    }
+    for (int i = column + cLen - 2; i >= column; i--) {
+      result.add(matrix[row + rLen - 1][i]);
+    }
+    for (int i = row + rLen - 2; i > row; i--) {
+      result.add(matrix[i][column]);
+    }
+    bfs(matrix, result, row + 1, column + 1, rLen - 2, cLen - 2);
+  }
+  // end::answer[]
+}