leetcode之Maximal Square

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 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 0

Return 4.

动态规划求解：

设dp[i][j]表示以matrix[i][j]结尾的最大正方形，则初始化：dp[i][j]=matrix[i][j],0<=i<=m,0<=j<=n(m,n为matrix的行数和列数)

状态转移方程：

dp[i][j] = MIN(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])+1,if matrix[i-1][j-1] == '1' and matrix[i][j] == '1' and matrix[i-1][j] == '1' and matrix[i][j-1] == '1'

代码：

class Solution {
#define MIN(a,b,c) (a)<=(b)?((a)<=(c)?(a):(c)):((b)<=(c)?(b):(c))
public:
int maximalSquare(vector<vector<char>>& matrix) {
if(matrix.empty())
{
return 0;
}
int m = matrix.size();
int n = matrix[0].size();
int dp[m+1][n+1];
int max = 0;
for(int i = 0;i < m;i ++)
{
for(int j = 0;j < n;j++)
{
if(matrix[i][j] == '1')
dp[i][j] = 1;
else
dp[i][j] = 0;
if(max < dp[i][j])
max = dp[i][j];
}
}
for(int i = 1;i < m;i ++)
{
for(int j = 1;j < n;j++)
{
if(matrix[i-1][j-1] == '1' && matrix[i][j] == '1' && matrix[i-1][j] == '1' && matrix[i][j-1] == '1')
{
int n = MIN(dp[i-1][j-1],dp[i-1][j],dp[i][j-1]);
dp[i][j] = n + 1;
}
else
{
}
if(max < dp[i][j])
max = dp[i][j];
}
}
return max * max;
}
};