一、题目
Given a rows x cols binary matrix filled with 0’s and 1’s, find the largest rectangle containing only 1’s and return its area.
Example 1:
Input: matrix = [[“1”,“0”,“1”,“0”,“0”],[“1”,“0”,“1”,“1”,“1”],[“1”,“1”,“1”,“1”,“1”],[“1”,“0”,“0”,“1”,“0”]]
Output: 6
Explanation: The maximal rectangle is shown in the above picture.
Example 2:
Input: matrix = [[“0”]]
Output: 0
Example 3:
Input: matrix = [[“1”]]
Output: 1
Constraints:
rows == matrix.length
cols == matrix[i].length
1 <= row, cols <= 200
matrix[i][j] is ‘0’ or ‘1’.
二、题解
class Solution {
public:
int maximalRectangle(vector<vector<char>>& matrix) {
int res = 0;
int m = matrix.size();
int n = matrix[0].size();
vector<int> heights(n,0);
for(int i = 0;i < m;i++){
for(int j = 0;j < n;j++){
heights[j] = matrix[i][j] == '0' ? 0 : heights[j] + 1;
}
res = max(res,largestRectangleArea(heights));
}
return res;
}
int largestRectangleArea(vector<int>& heights) {
int n = heights.size();
int res = 0;
stack<int> st;
for(int i = 0;i < n;i++){
while(!st.empty() && heights[i] <= heights[st.top()]){
int cur = st.top();
st.pop();
int left = st.empty() ? -1 : st.top();
res = max(res,(i - left - 1) * heights[cur]);
}
st.push(i);
}
while(!st.empty()){
int cur = st.top();
st.pop();
int left = st.empty() ? -1 : st.top();
res = max(res,(n - left - 1) * heights[cur]);
}
return res;
}
};
