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: return m, n = len(matrix), len(matrix[0]) length = 0 dp = [[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] = 1 else: 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
标签:最大,range,len,正方形,length,dp,matrix From: https://www.cnblogs.com/demo-deng/p/16632560.html