技術標籤：leetcode演算法leetcode資料結構棧

1 題目描述

給出一個只包含0和1的二維矩陣，找出最大的全部元素都是1的長方形區域，返回該區域的面積。

2 解題思路

https://blog.csdn.net/qq_41855420/article/details/87459549

3 程式碼實現

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        int n_rows = matrix.size();
        int n_cols =
 
 matrix[0].size();
        if(n_rows == 0 || n_cols == 0)
            return 0;
        
        vector<vector<int>> mat_trans(n_rows, vector<int>(n_cols));
        // 1st map: char2int
        for(int i = 0; i < n_rows; i++)
            for(int j = 0; j < n_cols; j++)
                mat_trans[
 
i][j] = matrix[i][j] - '0';
        // 2nd map: transform
        for(int i = 1; i < n_rows; i++)    // note that i = 1 at first
            for(int j = 0; j < n_cols; j++)
                if(mat_trans[i][j] == 1)
                    mat_trans[i][j] = mat_trans[i - 1][j] + 1;    // bravo!
        // 3rd map: monostack
 

        int max_area = 0;
        for(int i = 0; i < n_rows; i++){
            max_area = max(max_area, largestRectangleArea(mat_trans[i]));
        }
        return max_area;
    }
    
private:
    int largestRectangleArea(vector<int>& height) {
        // write code here
        int area = 0;
        stack<int> stk;
        height.push_back(0);
        for(int i = 0; i < height.size(); i++){
            if(stk.empty() || height[i] >= stk.top()){
                stk.push(height[i]);
            }
            else{
                int count = 0;
                while(!stk.empty() && height[i] < stk.top()){
                    count++;
                    area = max(area, count * stk.top());
                    stk.pop();
                }
                while(count--){
                    stk.push(height[i]);
                }
                stk.push(height[i]);
            }
        }
        return area;
    }
};

4 執行結果

執行時間：6ms
佔用記憶體：888k