1. 题目
给定一个仅包含 0 和 1 的二维二进制矩阵，找出只包含 1 的最大矩形，并返回其面积。
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
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximal-rectangle
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
2. DP解题
参考官方的解题思路：
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
}
};
例子的求解过程如下：
数组
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
left
[0 0 2 0 0]
[0 0 2 2 2]
[0 0 2 2 2]
[0 0 0 3 0]
right
[1 5 3 5 5]
[1 5 3 5 5]
[1 5 3 5 5]
[1 5 5 4 5]
height
[1 0 1 0 0]
[2 0 2 1 1]
[3 1 3 2 2]
[4 0 0 3 0]
area
[1 0 1 0 0]
[2 0 2 3 3]
[3 5 3 6 6]
[4 0 0 3 0]
