85. 最大矩形

 

难度困难1294

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。

 

示例 1：

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 = []
输出：0

示例 3：

输入：matrix = [["0"]]
输出：0

示例 4：

输入：matrix = [["1"]]
输出：1

示例 5：

输入：matrix = [["0","0"]]
输出：0

 

class Solution:
    def maximalRectangle(self, matrix: List[List[str]]) -> int:
 
        def max_a(heights):
            heights = [0] + heights + [0]
            res = 0
            stack = []
            for i in range(len(heights)):
                cur = heights[i]
                while stack and cur < heights[stack[-1]]:
                    top =stack.pop()
                    area = heights[top] * (i-stack[-1]-1)
                    res = max(res,area)
                stack.append(i)
            return res

        n1 = len(matrix)
        n2 = len(matrix[0])
        f_res = 0
        heights = [0] * n2
        for i in range(0,n1):
            for j in range(0,n2):
                if matrix[i][j] == '1':
                    heights[j]+=1
                else:
                    heights[j] = 0
            f_res = max(f_res,max_a(heights))
        return f_res

 

 

 

 

 

 

class Solution {
public:
    // ！！！！不要传引用，主函数中 每层都会对hights 修改
    int largestRectangleArea(vector<int> heights) {
        int res = 0;
        stack<int> stk;
        heights.insert(heights.begin(),0);
        heights.emplace_back(0);
        for(int i = 0; i < heights.size();i++) {
            while(!stk.empty() && heights[i] < heights[stk.top()]) {
                int top = stk.top();
                stk.pop();
                int w = i-stk.top()-1;
                int h = heights[top];
                res = max(res,w*h);
            }
            stk.push(i);
        }
        return res;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int res = 0;
        vector<int> heights = vector<int>(matrix[0].size(),0);
        for(int i = 0; i< matrix.size();i++) {
            for(int j = 0;j < matrix[0].size();j++){
                if (matrix[i][j]== '1') {
                    heights[j]+=1;
                } else {
                    heights[j] = 0;
                }
            }
            res = max(res,largestRectangleArea(heights));
        }
        return res;
    }
};

 

 

 

 

解法一 暴力破解
遍历每个点，求以这个点为矩阵右下角的所有矩阵面积。如下图的两个例子，橙色是当前遍历的点，然后虚线框圈出的矩阵是其中一个矩阵。

 

 

 

 

 

 

 

 

 

作者：windliang
链接：https://leetcode.cn/problems/maximal-rectangle/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-1-8/
来源：力扣（LeetCode）
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