二刷221

Author: diguage

docs/0221-maximal-square.adoc

@@ -1,54 +1,79 @@
 [#0221-maximal-square]
-= 221. Maximal Square
+= 221. 最大正方形
 
-{leetcode}/problems/maximal-square/[LeetCode - Maximal Square^]
+https://leetcode.cn/problems/maximal-square/[LeetCode - 221. 最大正方形 ^]
 
-Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
+在一个由 `0` 和 `1` 组成的二维矩阵内，找到只包含 `1` 的最大正方形，并返回其面积。
 
-.Example:
-----
-Input:
+*示例 1：*
 
-1 0 1 0 0
-1 0 1 1 1
-1 1 1 1 1
-1 0 0 1 0
+image::images/0221-01.jpg[{image_attr}]
 
-Output: 4
-----
+....
+输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
+输出：4
+....
 
-根据左上、左边、上边加自身，确定一个最小正方形；然后再进一步，在小正方形基础上，从四周选择已有正方形中最小，然后扩大，再这个过程中记录下最大的正方形边长，即可得到结果。思路如下图：
+*示例 2：*
 
-image::images/0221-1.png[{image_attr}]
+image::images/0221-02.jpg[{image_attr}]
 
-简化一下，不需要矩阵来存储所有正方形的计算结果，只需要一行来记录上一行计算结果和当前行计算结果即可。
+....
+输入：matrix = [["0","1"],["1","0"]]
+输出：1
+....
 
-image::images/0221-2.png[{image_attr}]
+*示例 3：*
 
-== 参考资料
+....
+输入：matrix = [["0"]]
+输出：0
+....
 
-. {leetcode}/problems/maximal-square/solution/[Maximal Square solution - LeetCode^]
+*提示：*
 
-Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
+* `m == matrix.length`
+* `n == matrix[i].length`
+* `+1 <= m, n <= 300+`
+* `matrix[i][j]` 为 `0` 或 `1`
 
-*Example:*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input: 
-*
-1 0 1 0 0
-1 0 <font color="red">1 <font color="red">1 1
-1 1 <font color="red">1 <font color="red">1 1
-1 0 0 1 0
-
-*Output:* 4
-----
+== 思路分析
+
+根据左上、左边、上边加自身，确定一个最小正方形；然后再进一步，在小正方形基础上，从四周选择已有正方形中最小，然后扩大，再这个过程中记录下最大的正方形边长，即可得到结果。思路如下图：
+
+image::images/0221-03.png[{image_attr}]
 
+简化一下，不需要矩阵来存储所有正方形的计算结果，只需要一行来记录上一行计算结果和当前行计算结果即可。
+
+image::images/0221-04.png[{image_attr}]
+
+无需另外开辟矩阵存储中间结果，直接在参数矩阵上存储即可。
+
+image::images/0221-05.png[{image_attr}]
 
 [[src-0221]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0221_MaximalSquare.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0221_MaximalSquare_2.java[tag=answer]
+----
+--
+====
+
+== 参考资料
 
+. https://leetcode.cn/problems/maximal-square/solutions/234964/zui-da-zheng-fang-xing-by-leetcode-solution/[221. 最大正方形 - 官方题解^]

docs/images/0221-01.jpg

@@ -6,7 +6,7 @@ endif::[]
 :doc_base_url: link:../docs
 
 
-[cols="7,36,7,50",options="header"]
+[cols="7,32,7,54",options="header"]
 |===
 |序号 |题目 |题解 |备注
 
@@ -335,6 +335,11 @@ endif::[]
 |{doc_base_url}/1289-minimum-falling-path-sum-ii.adoc[题解]
 |⭕️ 动态规划。题目理解错误。
 
+|{counter:codes2503}
+|{leetcode_base_url}/maximal-square/[221. 最大正方形^]
+|{doc_base_url}/0221-maximal-square.adoc[题解]
+|✅ 动态规划。直接将结果存储在参数矩阵上。如果正方形想扩大，则左边，左上和上面三个都是正方形时才可以，可以直接去这三者中的最小值。
+
 
 |===
 

docs/images/0221-02.jpg

@@ -18,6 +18,9 @@ public class _0221_MaximalSquare {
      * Memory Usage: 43.3 MB, less than 91.18% of Java online submissions for Maximal Square.
      *
      * Copy from: https://leetcode.com/problems/maximal-square/solution/[Maximal Square solution - LeetCode]
+     *
+     * @author D瓜哥 · https://www.diguage.com
+     * @since 2020-01-28 12:21
      */
     public int maximalSquare(char[][] matrix) {
         if (Objects.isNull(matrix) || matrix.length == 0) {

docs/images/0221-1.png renamed to docs/images/0221-03.png

@@ -0,0 +1,59 @@
+package com.diguage.algo.leetcode;
+
+public class _0221_MaximalSquare_2 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-04-18 23:35:51
+   */
+  public int maximalSquare(char[][] matrix) {
+    int row = matrix.length;
+    int col = matrix[0].length;
+    int result = 0;
+    for (int r = 0; r < row; r++) {
+      for (int c = 0; c < col; c++) {
+        if (matrix[r][c] == '0') {
+          continue;
+        }
+        // 加速
+        if (result == 0) {
+          result = 1;
+        }
+        char left = '0';
+        if (0 <= c - 1) {
+          left = matrix[r][c - 1];
+        }
+        // 加速
+        if (left == '0') {
+          continue;
+        }
+        char top = '0';
+        if (0 <= r - 1) {
+          top = matrix[r - 1][c];
+        }
+        // 加速
+        if (top == '0') {
+          continue;
+        }
+        char topLeft = matrix[r - 1][c - 1];
+        // 加速
+        if (topLeft == '0') {
+          continue;
+        }
+        matrix[r][c] = (char) (1 + Math.min(left, Math.min(topLeft, top)));
+        result = Math.max(result, matrix[r][c] - '0');
+      }
+    }
+    return result * result;
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0221_MaximalSquare_2().maximalSquare(new char[][]{
+      {'1', '0', '1', '0', '0'},
+      {'1', '0', '1', '1', '1'},
+      {'1', '1', '1', '1', '1'},
+      {'1', '0', '0', '1', '0'}
+    });
+  }
+}