[leetcode 85] Maximal Rectangle 待续

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最新推荐文章于 2025-09-05 23:16:08 发布

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分类专栏： [LeetCode 刷题] 文章标签： leetcode stack

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本文介绍了一种求解二维二进制矩阵中全由1构成的最大矩形面积的方法。该方法通过转换问题为求解直方图最大矩形面积，并使用堆栈辅助计算，最终实现了高效的解决方案。

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

思路：同[leetcode 84] Largest Rectangle in Histogram

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        const int m = matrix.size();
        if (m < 1) {
            return 0;
        }
        const int n = matrix[0].size();
        vector<vector<int> > intmatrix(m, vector<int>(n,0));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j <n; j++) {
                if (matrix[i][j] == '1') {
                    intmatrix[i][j] = 1;
                }
            }
        }
        for (int i = 1; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (intmatrix[i][j]>0) {
                    intmatrix[i][j] += intmatrix[i-1][j];
                }
            }
        }
        int res = 0;
        for (int i = 0; i < m; i++) {
            res = max(res, maxHis(intmatrix[i]));
        }
        return res;
        
    }
    int maxHis(vector<int> matrix) {
        int res = 0;
        matrix.push_back(0);
        int n = matrix.size();
        stack<int> p;
        for (int i = 0; i < n; ) {
            if (p.empty() || matrix[i] > matrix[p.top()]) {
                p.push(i++);
            } else {
                int tmp = p.top();
                p.pop();
                res = max(res, matrix[tmp]*(p.empty()?i:i-p.top()-1));
            }
        }
        return res;
    }
};