一刷85

diguage · diguage · commit d3889cebef73 · 2025-06-25T21:38:25.000+08:00
diff --git a/README.adoc b/README.adoc
@@ -621,12 +621,12 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 |Hard
 |
 
-//|{counter:codes}
-//|{leetcode_base_url}/maximal-rectangle/[85. Maximal Rectangle^]
-//|{source_base_url}/_0085_MaximalRectangle.java[Java]
-//|{doc_base_url}/0085-maximal-rectangle.adoc[题解]
-//|Hard
-//|
+|{counter:codes}
+|{leetcode_base_url}/maximal-rectangle/[85. Maximal Rectangle^]
+|{source_base_url}/_0085_MaximalRectangle.java[Java]
+|{doc_base_url}/0085-maximal-rectangle.adoc[题解]
+|Hard
+|
 
 |{counter:codes}
 |{leetcode_base_url}/partition-list/[86. Partition List^]
diff --git a/docs/0085-maximal-rectangle.adoc b/docs/0085-maximal-rectangle.adoc
@@ -1,28 +1,80 @@
 [#0085-maximal-rectangle]
-= 85. Maximal Rectangle
+= 85. 最大矩形
 
-{leetcode}/problems/maximal-rectangle/[LeetCode - Maximal Rectangle^]
+https://leetcode.cn/problems/maximal-rectangle/[LeetCode - 85. 最大矩形 ^]
 
-Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
+给定一个仅包含 `0` 和 `1` 、大小为 `rows x cols` 的二维二进制矩阵，找出只包含 `1` 的最大矩形，并返回其面积。
 
-*Example:*
+*示例 1：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:*
-[
-  ["1","0","1","0","0"],
-  ["1","0","*1*","*1*","*1*"],
-  ["1","1","*1*","*1*","*1*"],
-  ["1","0","0","1","0"]
-]
-*Output:* 6
-----
+image::images/0085-01.png[{image_attr}]
+
+....
+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：6
+解释：最大矩形如上图所示。
+....
+
+*示例 2：*
+
+....
+输入：matrix = [["0"]]
+输出：0
+....
+
+*示例 3：*
+
+....
+输入：matrix = [["1"]]
+输出：1
+....
+
+*提示：*
 
+* `rows == matrix.length`
+* `cols == matrix[0].length`
+* `+1 <= row, cols <= 200+`
+* `matrix[i][j]` 为 `0` 或 `1`
+
+
+== 思路分析
+
+还想暴力解决，奈何一团浆糊！
+
+计算每一行的高度，利用这个高度，使用与 xref:0084-largest-rectangle-in-histogram.adoc[84. 柱状图中最大的矩形] 中单调栈的思路来解决这个问题！
+
+image::images/0085-10.png[{image_attr}]
 
 [[src-0085]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0085_MaximalRectangle.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0085_MaximalRectangle_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
 
+. https://leetcode.cn/problems/maximal-rectangle/solutions/9535/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-1-8/[85. 最大矩形 - 详细通俗的思路分析，多解法^] -- 解决非常细致！
+. https://leetcode.cn/problems/maximal-rectangle/solutions/535672/zui-da-ju-xing-by-leetcode-solution-bjlu/[85. 最大矩形 - 官方题解^]
+. https://leetcode.cn/problems/maximal-rectangle/solutions/535898/python3-qian-zhui-he-dan-diao-zhan-ji-su-vkpp/[85. 最大矩形 - 前缀和+单调栈计算最大矩形^]
diff --git a/docs/images/0085-01.png b/docs/images/0085-01.png
diff --git a/docs/images/0085-10.png b/docs/images/0085-10.png
diff --git a/docs/index.adoc b/docs/index.adoc
@@ -253,7 +253,7 @@ include::0083-remove-duplicates-from-sorted-list.adoc[leveloffset=+1]
 
 include::0084-largest-rectangle-in-histogram.adoc[leveloffset=+1]
 
-// include::0085-maximal-rectangle.adoc[leveloffset=+1]
+include::0085-maximal-rectangle.adoc[leveloffset=+1]
 
 include::0086-partition-list.adoc[leveloffset=+1]
 
diff --git a/logbook/202503.adoc b/logbook/202503.adoc
@@ -1058,6 +1058,11 @@ endif::[]
 |{doc_base_url}/0044-wildcard-matching.adoc[题解]
 |❌ 动态规划。尝试双指针失败！完全没有想到可以用动态规划来解决这个问题！
 
+|{counter:codes2503}
+|{leetcode_base_url}/maximal-rectangle/[85. 最大矩形^]
+|{doc_base_url}/0085-maximal-rectangle.adoc[题解]
+|❌ 单调栈！计算每一行的高度，然后利用这个高度，使用与 {leetcode_base_url}/largest-rectangle-in-histogram/[84. 柱状图中最大的矩形^] 中相同的思路，结合单调栈的模式来解决问题。多种解法，还有动态规划的解法。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。
diff --git a/src/main/java/com/diguage/algo/leetcode/_0085_MaximalRectangle.java b/src/main/java/com/diguage/algo/leetcode/_0085_MaximalRectangle.java
@@ -0,0 +1,63 @@
+package com.diguage.algo.leetcode;
+
+import java.util.ArrayDeque;
+import java.util.Deque;
+
+public class _0085_MaximalRectangle {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-06-25 20:59:45
+   */
+  public int maximalRectangle(char[][] matrix) {
+    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
+      return 0;
+    }
+    int[] heights = new int[matrix[0].length];
+    int result = 0;
+    for (int row = 0; row < matrix.length; row++) {
+      // 计算出每一行的高度
+      for (int column = 0; column < matrix[row].length; column++) {
+        if (matrix[row][column] == '1') {
+          heights[column] += 1;
+        } else {
+          heights[column] = 0;
+        }
+      }
+      // 利用每一行的高度，去计算可以组成的最大矩阵面积
+      result = Math.max(result, largestRectangleArea(heights));
+    }
+    return result;
+  }
+
+  private int largestRectangleArea(int[] heights) {
+    int length = heights.length;
+    if (length == 1) {
+      return heights[0];
+    }
+    int[] newHeights = new int[length + 2];
+    // 两端两个哨兵
+    newHeights[0] = 0;
+    newHeights[length + 1] = 0;
+    System.arraycopy(heights, 0, newHeights, 1, length);
+    heights = newHeights;
+    Deque<Integer> stack = new ArrayDeque<>();
+    stack.push(heights[0]);
+    int result = 0;
+    for (int i = 1; i < newHeights.length; i++) {
+      while (heights[i] < heights[stack.peek()]) {
+        Integer left = stack.pop();
+        // 以弹出的栈顶元素为高，因为有可能一柱擎天就是最大的面积
+        // 另外，弹出一个元素后，如果当前高度还是比栈顶元素低，
+        //      那么下次再弹出时，就会一个更低的高度去计算从栈顶元素到 i-1 的矩阵了
+        int hight = heights[left];
+        // 因为 heights[i] 已经变低了，所以，得从 i-1 开始想前算
+        int width = (i - 1) - stack.peek();
+        result = Math.max(result, hight * width);
+      }
+      stack.push(i);
+    }
+    return result;
+  }
+  // end::answer[]
+}
