LeetCode 221 Maximal Square DP
作者：互联网
Given an m x n binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Solution
设 \(dp[i][j]\) 表示能到达位置 \((i,j)\) 的最大正方形边长。对于初始化，即 \(i=0 \text{ or } j=0\)，最大的边长就是该位置的数字；否则的话考虑转移：
\[dp[i][j] = 1+\min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1]) \]即由周围的来进行限制，只能选最小的边长
点击查看代码
class Solution {
private:
int dp[303][303];
int ans = 0;
public:
int maximalSquare(vector<vector<char>>& matrix) {
if(matrix.empty())return 0;
int n = matrix.size(), m = matrix[0].size();
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if((i==0)||(j==0)||(matrix[i][j]=='0')){
dp[i][j] = matrix[i][j]-'0';
}
else{
dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1;
}
ans = max(ans, dp[i][j] * dp[i][j]);
}
}
return ans;
}
};
标签：Maximal,Square,matrix,int,303,DP,边长,dp,size 来源： https://www.cnblogs.com/xinyu04/p/16260728.html