链接:https://leetcode.com/problems/maximal-rectangle/description/
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
The DP solution proceeds row by row, starting from the first row. Let the maximal rectangle area at row i and column j be computed by [right(i,j) - left(i,j)]*height(i,j).
All the 3 variables left, right, and height can be determined by the information from previous row, and also information from the current row. So it can be regarded as a DP solution. The transition equations are:
left(i,j) = max(left(i-1,j), cur_left), cur_left can be determined from the current row
right(i,j) = min(right(i-1,j), cur_right), cur_right can be determined from the current row
height(i,j) = height(i-1,j) + 1, if matrix[i][j]=='1';
height(i,j) = 0, if matrix[i][j]=='0'
class Solution {
public:
int maximalRectangle(vector<vector<char>>& matrix) {
if(matrix.empty() || matrix[0].empty())
return 0;
int m=matrix.size(),n=matrix[0].size(), res=0;
vector<int> left(n,0), right(n, n), height(n,0);
for(int i=0; i<m; i++)
{
int cur_left=0, cur_right=n;
for(int j=0; j<n; j++) // compute height (can do this from either side)
{
if(matrix[i][j]=='1')
height[j]++;
else
height[j]=0;
}
for(int j=0; j<n; j++) // compute left (from left to right)
{
if(matrix[i][j]=='1')
left[j]=max(left[j],cur_left);
else
{
left[j]=0;
cur_left=j+1;
}
}
for(int j=n-1; j>=0; j--) // compute right (from right to left)
{
if(matrix[i][j]=='1')
right[j]=min(right[j],cur_right);
else
{
right[j]=n;
cur_right=j;
}
}
for(int j=0; j<n; j++) // compute the area of rectangle (can do this from either side)
res = max(res,(right[j]-left[j])*height[j]);
}
return res;
}
};