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;}
};