LeetCode85. Maximal Rectangle——单调栈

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;
    }
};