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  • [Leetcode] Maximal Rectangle

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

    给定一个矩阵中,只有0和1,求出这个矩阵的一个最大的子矩阵,其中只包含1.

    例如

    01101

    11010

    01110

    11110

    11111

    00000

    其实这个问题可以转化为Largest Rectangle in Histogram,先将上面的矩阵转化为:

    01101

    12010

    03110

    14210

    25321

    00000

    然后对每一行求直方图的最大面积。

    class Solution {
    public:
        int maxArea(vector<int> &line) {
            if (line.size() < 1) return 0;
            stack<int> S;
            line.push_back(0);
            int sum = 0;
            for (int i = 0; i < line.size(); i++) {
                if (S.empty() || line[i] > line[S.top()]) S.push(i);
                else {
                    int tmp = S.top();
                    S.pop();
                    sum = max(sum, line[tmp]*(S.empty()? i : i-S.top()-1));
                    i--;
                }
            }
            return sum;
        }
        
        int maximalRectangle(vector<vector<char> > &matrix) {
            if (matrix.size() < 1) return 0;
            int n = matrix.size();
            if (n == 0) return 0;
            int m = matrix[0].size();
            if (m == 0) return 0;
            vector<vector<int> > lines(n, vector<int>(m, 0));
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    if (i == 0) {
                        lines[i][j] = ((matrix[i][j] == '1') ? 1 : 0);
                    } else {
                        lines[i][j] += ((matrix[i][j] == '1') ? lines[i-1][j] + 1 : 0);
                    }
                }
            }
            int maxRec = 0, tmpRec;
            for (int i = 0; i < n; ++i) {
                tmpRec = maxArea(lines[i]);
                maxRec = (maxRec > tmpRec) ? maxRec : tmpRec;
            }
            return maxRec;
        }
    };
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  • 原文地址:https://www.cnblogs.com/easonliu/p/3657489.html
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