feat: 语言优化

spiderlili · Apr 6, 2021 · 72e115d · 72e115d
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diff --git a/problems/84.largest-rectangle-in-histogram.md b/problems/84.largest-rectangle-in-histogram.md
@@ -11,7 +11,7 @@ https://leetcode-cn.com/problems/largest-rectangle-in-histogram/
 ![](https://tva1.sinaimg.cn/large/007S8ZIlly1ghltx8sr4uj305805odfn.jpg)
 
 以上是柱状图的示例，其中每个柱子的宽度为 1，给定的高度为  [2,1,5,6,2,3]。
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+
 ![](https://tva1.sinaimg.cn/large/007S8ZIlly1ghltx9kgd2j305805oa9z.jpg)
 
 图中阴影部分为所能勾勒出的最大矩形面积，其面积为  10  个单位。

diff --git a/problems/85.maximal-rectangle.md b/problems/85.maximal-rectangle.md
@@ -57,7 +57,7 @@ https://leetcode-cn.com/problems/maximal-rectangle/
 
 ## 代码
 
-代码支持：Python, CPP
+代码支持：Python
 
 Python Code:
 
@@ -88,42 +88,13 @@ class Solution:
 
 ```
 
-CPP Code:
-
-```cpp
-class Solution {
-public:
-    int maximalRectangle(vector<vector<char>>& A) {
-        if (A.empty() || A[0].empty()) return 0;
-        int ans = 0, M = A.size(), N = A[0].size();
-        vector<int> left(N, 0), right(N, N), height(N, 0);
-        for (int i = 0; i < M; ++i) {
-            int curLeft = 0, curRight = N;
-            for (int j = 0; j < N; ++j) height[j] = A[i][j] == '1' ? height[j] + 1 : 0;
-            for (int j = 0; j < N; ++j) {
-                if (A[i][j] == '1') left[j] = max(left[j], curLeft);
-                else {
-                    left[j] = 0;
-                    curLeft = j + 1;
-                }
-            }
-            for (int j = N - 1; j >= 0; --j) {
-                if (A[i][j] == '1') right[j] = min(right[j], curRight);
-                else {
-                    right[j] = N;
-                    curRight = j;
-                }
-            }
-            for (int j = 0; j < N; ++j) ans = max(ans, (right[j] - left[j]) * height[j]);
-        }
-        return ans;
-    }
-};
-```
-
 **复杂度分析**
 
 - 时间复杂度：$O(M * N)$
 - 空间复杂度：$O(N)$
 
 以上就是本文的全部内容了。大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库：https://github.com/azl397985856/leetcode 。 目前已经 38K star 啦。大家也可以关注我的公众号《力扣加加》带你啃下算法这块硬骨头。
+
+```
+
+```