LeetCode-85-Maximal Rectangle

最新推荐文章于 2025-02-23 17:11:32 发布

转载最新推荐文章于 2025-02-23 17:11:32 发布 · 76 阅读

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原文链接：http://www.cnblogs.com/nobodywang/p/10383983.html

本文介绍了一种解决二维二进制矩阵中寻找只包含1的最大矩形面积问题的算法。通过构建每行的直方图，并利用直方图最大面积的求解方法，实现了高效求解。关键步骤包括：对于矩阵中的每一行，根据当前行元素更新直方图高度，再求解直方图最大面积。

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算法描述：

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.

Example:

Input:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
Output: 6

解题思路：利用直方图最大面积求解，每一行构建一个直方图。当元素为1时，直方图对应元素加1，否则置为0；

 int maximalRectangle(vector<vector<char>>& matrix) {
        if(matrix.size()==0 || matrix[0].size()==0) return 0;
        int m = matrix.size();
        int n= matrix[0].size();
        int maxArea = 0;
        vector<int> hist(n,0);
        for(int i=0; i < m; i++){
            for(int j =0; j < n; j++){
                if( matrix[i][j] == '1')
                    hist[j] += 1;
                else
                    hist[j] = 0;
            }
            
            maxArea = max(maxArea,maxAreaOfHist(hist));
        }
        return maxArea;
       
    }

    int maxAreaOfHist(vector<int>& heights){
        if(heights.size()==0) return 0;
        stack<int> stk;
        int maxArea = 0;
        int i=0;
        while(i < heights.size()){
            if(stk.empty() || heights[i] >= heights[stk.top()]){
                stk.push(i);
                i++;
            }else{
                int index = stk.top();
                stk.pop();
                maxArea = max(maxArea, heights[index]*(stk.empty()? i: i-stk.top()-1));
            }

        }
        while(!stk.empty()){
            int index = stk.top();
            stk.pop();
            maxArea = max(maxArea, heights[index]*(stk.empty()?i:i-stk.top()-1));
        }
        return maxArea;
    }

转载于:https://www.cnblogs.com/nobodywang/p/10383983.html