题目链接:https://leetcode.com/problems/maximal-square/#/description
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
For example, given the following matrix:
1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0Return 4.
思路:
如果matrix[i][j]为1,那么A[i][j]=min(A[i-1][j-1],A[i-1][j],A[i][j-1])+1;如果matrix[i][j]为0,那么A[i][j]为0。
查看这个方程就好理解多了。
如下图:
最大的正方形边长为max{A[i][j]}。
此时时间复杂程度是O(M*N),空间复杂程度是O(M*N)。
class Solution{
public:
int maximalSquare(vector<vector<char>>& matrix)
{
int height=matrix.size();
if (height==0) return 0;
int width=matrix[0].size();
if(width==0) return 0;
vector<vector<int>> vec(height,vector<int>(width,0));
int res=0;
for(int i=0;i<height;i++)
{
for(int j=0;j<width;j++)
{
if(matrix[i][j]=='1')
{
vec[i][j]=1;
if(i>0 && j>0)
{
vec[i][j]+=min(min(vec[i-1][j],vec[i][j-1]),vec[i-1][j-1]);
}
res=max(res,vec[i][j]);
}
}
}
return res*res;
}
};