LeetCode开心刷题三十二天——85. Maximal Rectangle-蒲公英云

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

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:
https://www.twblogs.net/a/5bc0fa7d2b717711c9241544/zh-cn
The concept of monotone stack(单调栈) was introduced
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;
}
 python version:
1.python version why don't duplicate the C++'s thought,this
method looks more complex.
2.this version complex to understand but I add some print point help
understand now paste method
class Solution(object):
    def maximalRectangle(self, matrix):
        """
        :type matrix: List[List[str]]
        :rtype: int
        """
        if not matrix or not matrix[0]: return 0
        M, N = len(matrix), len(matrix[0])
        height = [0] * N
        res = 0
        for row in matrix:
            for i in range(N):
                if row[i] == '0':
                    height[i] = 0
                else:
                    height[i] += 1
                print("height1")
                print(height)
            res = max(res, self.maxRectangleArea(height))
        return res
    def maxRectangleArea(self, height):
        if not height: return 0
        print("height")
        print(height)
        res = 0
        stack = list()
        print("stack0")
        print(stack)
        height.append(0)
        print("height2")
        print(height)
        for i in range(len(height)):
            cur = height[i]
            while stack and cur < height[stack[-1]]:
                print("stack")
                print(stack)
                w = height[stack.pop()]
                h = i if not stack else i - stack[-1] - 1
                res = max(res, w * h)
            stack.append(i)
        return res
solu = Solution()
matirx=[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
print(solu.maximalRectangle(matirx))