一、内容
在一个由 '0' 和 '1' 组成的二维矩阵内，找到只包含 '1' 的最大正方形，并返回其面积。
示例 1：
输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：4
示例 2：
输入：matrix = [["0","1"],["1","0"]]
输出：1
示例 3：
输入：matrix = [["0"]]
输出：0
提示：
m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[i][j] 为 '0' 或 '1'
二、思路
- dp[i][j]代表第i行第j列能够组成的矩阵的最大长度
- 它有3个状态转移过来，dp[i-1][j-1], dp[i-1][j],dp[i][j-1]，这3个位置中最小长度+1便是新的正方形的最长长度
三、代码
class Solution {
public:
int maximalSquare(vector>& matrix) {
int n = matrix.size(), m = matrix[0].size(), ans = 0;
vector> dp(n + 1, vector(m + 1));
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (matrix[i - 1][j - 1] == '0') continue;
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]);
}
}
return ans * ans;
}
};
