LeetCode 85. 最大矩形（DP/单调递增栈，难）

示例:
输入:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
输出: 6

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& mat) {
        if(mat.empty())
            return 0;
        int i, j, minL, maxR, maxarea = 0;
        int r = mat.size(), c = mat[0].size();
        vector<vector<int>> left(r,vector<int>(c,0));
        vector<vector<int>> right(r,vector<int>(c,c));
        vector<vector<int>> height(r,vector<int>(c,0));
        for(i = 0; i < r; i++) 
        {
            //填写left，相连的1,先到最高，然后最左侧的下标
            minL = 0;
            for(j = 1; j < c; j++)
            {
                if(i == 0)//第一行
                {
                    if(mat[i][j] == '1')
                    {
                        if(mat[i][j-1] == '0')
                            minL = j;//左边0，当前1，需要更新最左边的边界minL
                        left[i][j] = minL;
                    }
                }
                else//剩余行
                {
                    if(mat[i][j] == '1')
                    {
                        if(mat[i][j-1] == '0')
                            minL = j;
                        left[i][j] = max(minL,left[i-1][j]);//跟上面的行，比较，取大
                    }
                }
            }

            maxR = c;
            for(j = c-2; j >= 0; j--)
            {
                if(i == 0)//第一行
                {
                    if(mat[i][j] == '1')
                    {
                        if(mat[i][j+1] == '0')
                            maxR = j+1;//右边0，当前1，更新最右边的边界maxR
                        right[i][j] = maxR;
                    }
                }
                else//其余
                {
                    if(mat[i][j] == '1')
                    {
                        if(mat[i][j+1] == '0')
                            maxR = j+1;
                        right[i][j] = min(maxR,right[i-1][j]);//还要更上面的比较，取小
                    }
                }
            }

            for(j = 0; j < c; j++)
            {
                if(i == 0)//第一行
                {
                    if(mat[i][j] == '1')
                        height[i][j] = 1;
                }
                else//剩余
                {
                    if(mat[i][j] == '1')
                        height[i][j] = 1+height[i-1][j];
                }
            }

            for(j = 0; j < c; j++)
                maxarea = max(maxarea, (right[i][j]-left[i][j])*height[i][j]);
        }
        return maxarea;//返回最大面积
    }
};

  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]

  [0  0  2  0  0]
  [0  0  2  2  2]
  [0  0  2  2  2]
  [0  0  0  3  0]

  [1  5  3  5  5]
  [1  5  3  5  5]
  [1  5  3  5  5]
  [1  5  5  4  5]

  [1  0  1  0  0]
  [2  0  2  1  1]
  [3  1  3  2  2]
  [4  0  0  3  0]

  [1  0  1  0  0]
  [2  0  2  3  3]
  [3  5  3  6  6]
  [4  0  0  3  0]

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& mat) {
        if(mat.empty())
            return 0;
        int i, j, hi, width, maxarea = 0, m = mat.size(), n = mat[0].size();
        vector<int> h(n+1, 0);
        for(i = 0; i < m; ++i)
        {
            stack<int> s;
            mat[i].push_back('0');//请看84题
            for(j = 0; j <= n; ++j)
            {
                h[j] = mat[i][j]=='1' ? h[j]+1 : 0;//根据前一行，得到当前行的高
                while(!s.empty() && h[s.top()] > h[j])
                {
                    hi = h[s.top()];
                    s.pop();
                    width = s.empty() ? j : j-s.top()-1;
                    maxarea = max(maxarea, hi*width);
                }
                s.push(j);
            }
        }
        return maxarea;
    }
};