LeetCode-Maximal Rectangle

看清问题，用DP就可以了，

之前我首先想到的是图的搜索，实际上在这里并不适用；

必须遍历每个点，计算它的最大矩阵，

遍历的顺序为从左上角到右下角，

x, y表示左边的最大长度，y表示上面的最大长度，

然后计算每个点的最大面积；

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int res = 0;
        if (matrix.empty() || matrix[0].empty())
            return res;
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<pair<int, int> > > dp(m, vector<pair<int, int> >(n));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == '0')
                    continue;
                int x = (j == 0) ? 1 : dp[i][j - 1].first + 1;
                int y = (i == 0) ? 1 : dp[i - 1][j].second + 1;
                dp[i][j] = make_pair(x, y);
                int height = y;
                for (int k = j; k > j - x; --k) {
                    height = min(height, dp[i][k].second);
                    res = max(res, height * (j - k + 1));
                }
            }
        }
        return res;
    }
};

也可以先计算每一列的节点的最大长度，然后遍历该列所有点，

计算出最大面积，这个过程可以借用栈来完成；

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int res = 0;
        if (matrix.empty() || matrix[0].empty())
            return res;
        int m = matrix.size();
        int n = matrix[0].size();
        vector<int> height(n + 1, 0);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                height[j] = (matrix[i][j] == '0') ? 0: height[j] + 1;
            }
            res = max(res, getRect(height));
        }
        return res;
    }
    int getRect(const vector<int> height) {
        stack<int> st;
        int res = 0, m = height.size();
        for (int i = 0; i < m; ++i) {
            int count = 0;
            while (!st.empty() && st.top() > height[i]) {
                ++count;
                res = max(res, count * st.top());
                st.pop();
            }
            while (count--)
                st.push(height[i]);
            st.push(height[i]);
        }
        return res;
    }
};