问题描述 链接到标题
85. 最大矩形 (Hard)
给定一个仅包含 0
和 1
、大小为 rows x cols
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 1:

输入:matrix = [["1","0","1","0","0"],["1","0","1","1","1"],
["1","1","1","1","1"],["1","0","0","1","0"]]
输出:6
解释:最大矩形如上图所示。
示例 2:
输入:matrix = []
输出:0
示例 3:
输入:matrix = [["0"]]
输出:0
示例 4:
输入:matrix = [["1"]]
输出:1
示例 5:
输入:matrix = [["0","0"]]
输出:0
提示:
rows == matrix.length
cols == matrix[0].length
1 <= row, cols <= 200
matrix[i][j]
为'0'
或'1'
解题思路 链接到标题
其实本题就相当于是 84. 柱状图中最大的矩形 (Hard) 套了一层皮,想到了这一点之后就不难了。
代码 链接到标题
class Solution {
public:
int Count(int row, int n, vector<int> &vec, vector<vector<char>> &matrix) {
for (int i = 0; i < n; ++i) {
vec[i] = matrix[row][i] == '0' ? 0 : vec[i] + 1;
}
// 单调递增栈
stack<int> stk;
int res = 0;
for (int i = 0; i < n; ++i) {
while (!stk.empty() && vec[i] < vec[stk.top()]) {
int idx = stk.top(), h = vec[idx];
stk.pop();
int rec = stk.empty() ? h * i : h * (i - stk.top() - 1);
res = max(rec, res);
}
stk.push(i);
}
while (!stk.empty()) {
int idx = stk.top();
int h = vec[idx];
stk.pop();
int rec = stk.empty() ? h * n : h * (n - stk.top() - 1);
res = max(res, rec);
}
return res;
}
int maximalRectangle(vector<vector<char>> &matrix) {
// 二维前缀和?
// 转化为 84 题,即可用单调栈解决
int m = matrix.size(), n = matrix[0].size();
int res = 0;
vector<int> vec(n);
for (int i = 0; i < m; ++i) {
res = max(res, Count(i, n, vec, matrix));
}
return res;
}
};