leetcode 85. Maximal Rectangle

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

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 6.

我曹，还是单调栈。

记录下以当前行向上能得到的连续的1的个数（比如有sum[i]个），统计以这个sum[i]为最小值，向左，向右能扩展的最大距离。

class Solution {
public:
    class node {
        public:
        int x, left, right;
    };
    int solve(int n, int m, vector<int>& sum) {
        vector<node> v;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            if (v.empty()) v.push_back({sum[i], i, i});
            else {
                node u = {sum[i], i, i};
                while(!v.empty() && sum[i] < v.back().x) {
                    node tmp = v.back(); v.pop_back();
                    if (!v.empty()) v.back().right = tmp.right;
                    u.left = tmp.left;
                    int y = (tmp.right - tmp.left + 1) * tmp.x;
                    if (y > ans) ans = y;
                }
                v.push_back(u);
            }
        }
        while (!v.empty()) {
            node tmp = v.back();
            v.pop_back();
            int y = (tmp.right - tmp.left + 1) * tmp.x;
            ans = max(ans, y);
            if (!v.empty()) v.back().right = tmp.right;
        }
        return ans;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n = matrix.size();
        if (n == 0) return 0;
        int m = matrix[n-1].size();
        vector<int> sum(m);
        int ans = 0;
        for (int i = 0; i < m; ++i) sum[i] = (matrix[0][i] - '0');
        ans = solve(n, m, sum);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (matrix[i][j] == '1') sum[j]++;
                else sum[j] = 0;
            }
            ans = max(ans, solve(n, m, sum));
        }
        return ans;
    }
};