78 lines
2.4 KiB
Markdown
78 lines
2.4 KiB
Markdown
## 题目地址（85. 最大矩形）
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https://leetcode-cn.com/problems/maximal-rectangle/
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## 题目描述
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给定一个仅包含 0 和 1 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。
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示例：
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输入：
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```
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[
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["1","0","1","0","0"],
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["1","0","1","1","1"],
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["1","1","1","1","1"],
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["1","0","0","1","0"]
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]
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```
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输出：6
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## 思路
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我在 [【84. 柱状图中最大的矩形】多种方法（Python3）](https://leetcode-cn.com/problems/largest-rectangle-in-histogram/solution/84-zhu-zhuang-tu-zhong-zui-da-de-ju-xing-duo-chong/ "【84. 柱状图中最大的矩形】多种方法（Python3）") 使用了多种方法来解决。 然而在这道题，我们仍然可以使用完全一样的思路去完成。 不熟悉的可以看下我的题解。本题解是基于那道题的题解来进行的。
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拿题目给的例子来说：
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```
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[
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["1","0","1","0","0"],
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["1","0","1","1","1"],
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["1","1","1","1","1"],
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["1","0","0","1","0"]
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]
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```
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我们逐行扫描得到 `84. 柱状图中最大的矩形` 中的 heights 数组：
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这样我们就可以使用`84. 柱状图中最大的矩形` 中的解法来进行了，这里我们使用单调栈来解。
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## 代码
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```python
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class Solution:
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def largestRectangleArea(self, heights: List[int]) -> int:
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n, heights, st, ans = len(heights), [0] + heights + [0], [], 0
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for i in range(n + 2):
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while st and heights[st[-1]] > heights[i]:
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ans = max(ans, heights[st.pop(-1)] * (i - st[-1] - 1))
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st.append(i)
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return ans
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def maximalRectangle(self, matrix: List[List[str]]) -> int:
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m = len(matrix)
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if m == 0: return 0
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n = len(matrix[0])
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heights = [0] * n
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ans = 0
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for i in range(m):
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for j in range(n):
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if matrix[i][j] == "0":
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heights[j] = 0
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else:
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heights[j] += 1
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ans = max(ans, self.largestRectangleArea(heights))
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return ans
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```
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**复杂度分析**
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- 时间复杂度：$O(M * N)$
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- 空间复杂度：$O(N)$
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欢迎关注我的公众号《脑洞前端》获取更多更新鲜的 LeetCode 题解
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