Given a 2D binary matrix filled with 0's and 1's, find the largest square 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: 4
思路:
看到这道题首先想到的就是之前的求解最大矩阵的一道题(第85题),但是这道有更加优化的做法,我们使用动态规划的思想来完成该题,解决方案如下:
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.size() == 0) return 0;
int row = int(matrix.size());
int col = int(matrix[0].size());
vector<vector<int>> dp(row, vector<int>(col, 0));
int maxEdge = 0;
for(int i = 0;i<row;i++)
if(matrix[i][0] == '1') {dp[i][0] = 1; maxEdge = 1;}
for(int j = 1;j<col;j++)
if(matrix[0][j] == '1') {dp[0][j] = 1; maxEdge = 1;}
for (int i = 1; i < row; i++) {
for (int j = 1; j < col; j++) {
if (matrix[i][j] == '0'){
dp[i][j] = 0;
continue;
}
vector<int> candidate; //in case of no element
candidate.push_back(dp[i - 1][j]);
candidate.push_back(dp[i][j - 1]);
candidate.push_back(dp[i - 1][j - 1]);
dp[i][j] = *min_element(candidate.begin(), candidate.end()) + 1;
maxEdge = max(maxEdge, dp[i][j]);
}
}
return maxEdge*maxEdge;
}
};