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
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2. 解题

类似题目：
LeetCode 221. 最大正方形（DP）
LeetCode 84. 柱状图中最大的矩形（单调递增栈）

2.1 DP

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minLleft[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maxRright[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]

2.2 单调递增栈

• 思路跟84题一致，行数变多了而已

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;}
};