LEETCODE: Maximal Rectangle

最新推荐文章于 2025-12-05 14:37:41 发布

原创最新推荐文章于 2025-12-05 14:37:41 发布 · 236 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LEETCODE #算法

LEETCODE 专栏收录该内容

153 篇文章

订阅专栏

本文介绍了一种算法，用于在二维二进制矩阵中找到面积最大的全一矩形，并返回其面积。

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

其实方法和（http://blog.youkuaiyun.com/anyicheng2015/article/details/42191979）类似。

class Solution {
public:
    int maxArea(vector<int> &line) {
        if(line.size() < 1) return 0;
        stack<int> exstack;
        line.push_back(0);
        int sum = 0;
        for(int ii = 0; ii < line.size(); ii ++) {
            if(exstack.empty() || line[ii] > line[exstack.top()]) {
                exstack.push(ii);
            }
            else{
                int temp = exstack.top();
                exstack.pop();
                sum = max(sum, line[temp] * (exstack.empty() ? ii : ii - exstack.top() - 1));
                ii --;
            }
        }
        return sum;
    }

    int maximalRectangle(vector<vector<char> > &matrix) {
        if(matrix.size() == 0) return 0;
        int n = matrix.size();
        int m = matrix[0].size();
        if(m == 0) return 0;
        vector<vector<int> > lines(n, vector<int>(m, 0));
        for(int ii = 0; ii < n; ii ++) {
            for(int jj = 0; jj < m; jj ++) {
                if(ii == 0) {
                    lines[ii][jj] = matrix[ii][jj] == '1' ? 1 : 0;
                }
                else {
                    lines[ii][jj] = matrix[ii][jj] == '1' ? lines[ii - 1][jj] + 1 : 0;
                }
            }
        }
        int maxval = 0;
        for(int ii = 0; ii < n; ii ++) {
            int newmaxval = maxArea(lines[ii]);
            maxval = (maxval > newmaxval) ? maxval : newmaxval;
        }
        return maxval;
    }
};