Leetcode 85. 最大矩形

借助上一题的答案，

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int Len = heights.size();
        vector<int> sta, L(Len, 0), R(Len, 0);
        for (int i = 0; i < Len; ++i) {
            while (!sta.empty() && heights[sta.back()] > heights[i]) {
                int x = sta.back(); sta.pop_back();
                R[x] = i - 1;
            }
            sta.push_back(i);
        }
        while (!sta.empty()) {
            int x = sta.back(); sta.pop_back();
            R[x] = Len - 1;
        }
        for (int i = Len - 1; i >= 0; --i) {
            while (!sta.empty() && heights[sta.back()] > heights[i]) {
                int x = sta.back(); sta.pop_back();
                L[x] = i + 1;
            }
            sta.push_back(i);
        }
        while (!sta.empty()) {
            int x = sta.back(); sta.pop_back();
            L[x] = 0;
        }
        int ans = 0;
        for (int i = 0; i < Len; ++i)
            ans = max(ans, (R[i] - L[i] + 1)*heights[i]);
        return ans;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n = matrix.size(), m = n == 0 ? 0 : matrix[0].size(), ans = 0;
        vector<vector<int>> up(n, vector<int>(m, 0)), left(n, vector<int>(m, 0));
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                if (matrix[i][j] == '1') {
                    if (i == 0) up[i][j] = 1;
                    else up[i][j] = up[i - 1][j] + 1;
                    if (j == 0) left[i][j] = 1;
                    else left[i][j] = left[i][j - 1] + 1;
                }
        vector<int> heights(m, 0);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j)
                heights[j] = up[i][j];
            ans = max(ans, largestRectangleArea(heights));
        }
        heights.assign(n, 0);
        for (int j = 0; j < m; ++j) {
            for (int i = 0; i < n; ++i)
                heights[i] = left[i][j];
            ans = max(ans, largestRectangleArea(heights));
        }
        return ans;
    }
};