问题：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。
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
解法：动态规划
注意事项：dp的递推公式及边界条件
class Solution:def maximalSquare(self, matrix: List[List[str]]) -> int:if not matrix or len(matrix) < 1:returnm, n = len(matrix), len(matrix[0])length = 0dp = [[0] * n for _ in range(m)] # dp[i][j]表示以(i,j)为右下角正方形的边长for i in range(m):for j in range(n):if matrix[i][j] == "1":if i==0 or j==0:dp[i][j] = 1else:dp[i][j] = min(dp[i-1][j], dp[i-1][j-1], dp[i][j-1]) + 1 # 通过最小边长确定不满足条件length = max(dp[i][j], length)return length*length
