难度: Medium
原题连接
内容描述
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 1 - 时间复杂度: O(row * col)- 空间复杂度: O(row * col)******
dp[i][j]代表以matrix[i][j]为右下角的正方形的最大长度
状态方程dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1,但是必须要满足dp[i][j] == 1,因为右下角也必须为1啊
原本的matrix DP
1 0 1 0 0 1 0 1 0 0
1 0 1 1 1 → 1 0 1 1 1
1 1 1 1 1 1 1 1 2 2
1 0 0 1 0 1 0 0 1 0
beats 48.42%
class Solution(object):
def maximalSquare(self, matrix):
"""
:type matrix: List[List[str]]
:rtype: int
"""
row = len(matrix)
col = len(matrix[0]) if row else 0
dp = [[int(matrix[i][j]) for j in range(col)] for i in range(row)]
for i in range(1, row):
for j in range(1, col):
if dp[i][j] == 1:
dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1
max_len = 0
for i in range(row):
for j in range(col):
max_len = max(max_len, dp[i][j])
return max_len * max_len