https://leetcode.com/problems/maximal-square/#/description
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest square containing only 1‘s and return its area.
For example, given the following matrix:
1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0
Return 4.
Sol 1:
Brute Force.
class Solution(object):
def maximalSquare(self, matrix):
"""
:type matrix: List[List[str]]
:rtype: int
"""
# brute force
if len(matrix) <= 0 :
return 0
rows = len(matrix)
cols = len(matrix[0])
maxsqlen = 0
for i in range(rows):
for j in range(cols):
# initilize flag to be True
if matrix[i][j] == ‘1‘:
sqlen = 1
flag = True
# border limits
while sqlen + i < rows and sqlen + j < cols and flag:
# advance square length downwords
for k in range(j, sqlen + j + 1):
if matrix[i+sqlen][k] == ‘0‘:
flag = False
break
# advance square length to the right
for k in range(i, sqlen + i + 1):
if matrix[k][j+sqlen] == ‘0‘:
flag = False
break
if flag:
sqlen += 1
if maxsqlen < sqlen:
maxsqlen = sqlen
return maxsqlen * maxsqlen
Complexity Analysis
- Time complexity : O\big((mn)^2\big)O((mn)?2??). In worst case, we need to traverse the complete matrix for every 1.
- Space complexity : O(1)O(1). No extra space is used.
Sol 2:
DP.
（sth. wrong with initlization of DP...）
class Solution(object):
def maximalSquare(self, matrix):
"""
:type matrix: List[List[str]]
:rtype: int
"""
# DP
# Time complexity : O(mn)O(mn). Single pass.
# pace complexity : O(mn)O(mn). Another matrix of same size is used for dp.
if len(matrix) <= 0 :
return 0
rows = len(matrix)
cols = len(matrix[0])
maxsqlen = 0
dp = [‘0‘] * (rows + 1) * (cols + 1)
for i in range(1, rows+1):
for j in range(1, cols+1):
if matrix[i-1][j-1] == ‘1‘:
dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1
maxsqlen = max(maxsqlen, dp[i][j])
return maxsqlen * maxsqlen
时间： 2024-08-01 10:33:01