LeetCode开心刷题三十二天——85. Maximal Rectangle

Posted 2022-07-02 marigolci

85. Maximal Rectangle

Hard

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Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest rectangle containing only 1‘s and return its area.

Example:

Input:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
Output: 6

IDEA:
ONE IMPORTANT THING:
dp[i][j] means the number of "1" in i rows before j columns.
dp[1][2]代表在第一行中的第二列之前1的个数。

#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#include<climits>
using namespace std;
class Solution 
public:
    int maximalRectangle(vector<vector<char>>& matrix) 
        //NULL judge very important
        int n1=matrix.size();
        if(n1==0) return 0;
        int n2=matrix[0].size();
        //dp[i][j] is the max number of 1 in columns j rows i
        //vector initializer special no need ==
        vector<vector<int>> dp(n1,vector<int>(n2));
        //initialization
        for(int i=0;i<n1;i++)
        
            for(int j=0;j<n2;j++)
            
                //nested function need to pay attention whether finish all procedure such as this finish one easy to forget the latter ?:
                dp[i][j]=(matrix[i][j]==‘1‘)?(j==0?1:dp[i][j-1]+1):0;
            
        

        int ans=0;

        for(int i=0;i<n1;i++)
        
            for(int j=0;j<n2;j++)
            
                int len = INT_MAX;
                for(int k=i;k<n1;k++)
                
                    len=min(len,dp[k][j]);
                    if(len==0) break;
                    ans=max((k-i+1)*len,ans);
                
            
        

        return ans;
    
;

int main()

    vector<vector<char>> matrix=
    
        ‘1‘,‘0‘,‘1‘,‘0‘,‘0‘,
        ‘1‘,‘0‘,‘1‘,‘1‘,‘1‘,
        ‘1‘,‘1‘,‘1‘,‘1‘,‘1‘,
        ‘1‘,‘0‘,‘0‘,‘1‘,‘0‘
    ;
    Solution s;
    int res=s.maximalRectangle(matrix);
    cout<<res<<endl;
    return 0;