题目: 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。
来源: https://leetcode-cn.com/problems/maximal-square/
法一: 自己的代码
思路: 实际上是暴力解法,先记录每个位置向上和向左的可能的正方形边长,在逐个判断是否能够构成正方形.
注意这里要判断观察出正方形的最大边长只与上边,左边,左上的格子中最大正方形的边长有关.不止与上边和左边有关!写出动态转移方程是关键
# 执行用时 :264 ms, 在所有 Python3 提交中击败了35.46% 的用户 # 内存消耗 :15.7 MB, 在所有 Python3 提交中击败了5.32%的用户 from typing import List import math class Solution: def maximalSquare(self, matrix: List[List[str]]) -> int: # print(matrix) r = len(matrix) if r == 0: return 0 c = len(matrix[0]) memo = [[0] * c for i in range(r)] m = 0 if matrix[0][0] == ‘1‘: memo[0][0] = (1,1,1) m = 1 else: memo[0][0] = (0,0,0) # 先处理第一行和第一列 for i in range(1,r,1): if matrix[i][0] == ‘1‘: memo[i][0] = (1,1,1+memo[i-1][0][2]) m = 1 else: memo[i][0] = (0,0,0) for j in range(1,c,1): if matrix[0][j] == ‘1‘: memo[0][j] = (1,1+memo[0][j-1][1],1) m = 1 else: memo[0][j] = (0,0,0) # 判断其余的位置 for i in range(1,r,1): for j in range(1,c,1): if matrix[i][j] == ‘1‘: memo[i][j] = (1,1+memo[i][j-1][1],1+memo[i-1][j][2]) else: memo[i][j] = (0,0,0) # 判断是否在斜线上是否能够构成正方形,如果不能返回最小的正方形边长 def judge(k,i,j): r = min(k[1:]) cou = 1 while cou < r: if min(memo[i-cou][j][1],memo[i][j-cou][2]) >= cou+1: pass else: return cou cou += 1 return r for i in range(1,r,1): for j in range(1,c,1): if memo[i][j][0] == 0: continue elif min(memo[i][j][1],memo[i][j][2]) > math.sqrt(m): kkk = judge(memo[i][j], i, j) if kkk > math.sqrt(m): m = kkk ** 2 return m
法二: 别人的代码
思路: 利用备忘录,先对数组加两个0边(注意这里在原list上加边不方便,所以要在备忘录上加边),方便后序的动态规划.要学会这个技巧!
from typing import List class Solution: def maximalSquare(self, matrix: List[List[str]]) -> int: if(not matrix): return 0 m=len(matrix) n=len(matrix[0]) res=0 # dp作为备忘录,记录每个位置的最大正方形的边长 # 由于在原来的list中加边不方便,在dp中加边. dp=[[0]*(n+1) for _ in range(m+1)] for i in range(1,m+1): for j in range(1,n+1): if(matrix[i-1][j-1]=="1"): # 动态转移方程. dp[i][j]=min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])+1 res=max(dp[i][j],res) return res*res if __name__ == ‘__main__‘: duixiang = Solution() # a = duixiang.maximalSquare([["1","0","1","0","0"], # ["1","0","1","1","1"], # ["1","1","1","0","1"], # ["1","0","1","1","1"]] # ) a = duixiang.maximalSquare([["0","1","1","0","1"], ["1","1","0","1","0"], ["0","1","1","1","0"], ["1","1","1","1","0"], ["1","1","1","1","1"], ["0","0","0","0","0"]] ) print(a)
ttt
原文地址:https://www.cnblogs.com/xxswkl/p/12142020.html
时间: 2024-09-30 08:41:58